This provides a comprehensive guide on solving numerical problems related to electricity, including electric current, potential difference, Ohm's law, resistance, and resistivity. It presents various formulas and examples to calculate charge, current, voltage, and resistance in different scenarios. Additionally, it covers the combination of resistances in series and parallel configurations.
Formulas to Remember
To solve the numericals on electricity, we will use the following formulas:
$$ I = \frac{Q}{t} $$
$$ I = \frac{ne}{t} $$
Where:
I = Electric current – Ampere
Q = charge – Coulomb
T = time - second
N = number of electrons
E = charge on an electron = \( 1.6 \times 10^{-19} \)
Time (t) = 2.5 Hrs = \( 2.5 \times 60 \times 60 = 9000 \) sec
Charge (Q) = ?
$$ I = \frac{Q}{t} $$
$$ Q = It $$
$$ Q = 0.5 \times 9000 $$
Q = 4500 Coulomb
Charge (Q) = 10 C
Electric current (I) = ?
$$ I = \frac{Q}{t} $$
$$ I = \frac{10}{5} $$
I = 2 Amp.
Time (t) = 15 min = \( 15 \times 60 = 900 \) sec
Charge (Q) = ?
$$ I = \frac{Q}{t} $$
$$ Q = I \times t $$
$$ Q = 0.5 \times 900 $$
Q = 450 Coulomb
Time (t) = 15 s
Number of electrons (n) = ?
Charge on electrons (e) = \( 1.6 \times 10^{-19} \) C
$$ I = \frac{ne}{t} $$
After cross multiplication, we find:
$$ n = \frac{I \times t}{e} $$
Now put the given values in the formula:
$$ n = \frac{1 \times 10}{1.6 \times 10^{-19}} $$
$$ n = \frac{100}{16 \times 10^{-19}} $$
$$ n = \frac{6.25}{10^{-19}} $$
\( n = 6.25 \times 10^{19} \)
Number of electrons (n) has no unit required.
Charge (Q) = 150000 C
Charge required for 1 mole copper = 150000 C
Charge required for 0.2 mole copper = \( 150000 \times 0.2 = 30000 \) C
$$ I = \frac{Q}{t} $$
$$ t = \frac{Q}{I} = \frac{30000}{5} $$
t = 6000 S
Time (t) = 5 min = \( 5 \times 60 = 300 \) s
Charge (Q) = ?
$$ I = \frac{Q}{t} $$
$$ Q = I \times t $$
$$ Q = 10 \times 300 $$
Q = 3000 C
Time (t) = 10 s
Electric current (I) = ?
$$ I = \frac{ne}{t} $$
[e = \( 1.6 \times 10^{-19} \)]
$$ I = \frac{1.25 \times 10^{18} \times 1.6 \times 10^{-19}}{10} $$
I = 0.2 A
Number of electrons (n) = ?
Number of elections in 1 Coulombs is = \( 6.25 \times 10^{18} \)
Number of electrons in 0.1 Coulomb charge is = \( 6.25 \times 10^{18} \times 0.1 \)
= \( 6.25 \times 10^{17} \)
Formulas to Remember
$$ V = \frac{W}{Q} $$
Where:
V = Potential difference – volt
W = work done - Joule
Q = charge – Coulomb
Potential (V1) = 150 volt
Potential (V2) = 200 volt
Potential difference (V) = V2 - V1
200 – 150 = 50 volt
Work (W) = ?
$$ V = \frac{W}{Q} $$
$$ W = VQ $$
$$ W = 50 \times 8 = 400 J $$
Charge (Q) = 1 C
Energy (E) = ?
$$ V = \frac{W}{Q} $$
Work is defined as energy transferred by force, so We can put E in the place of W
$$ E = VQ $$
= 10 × 1 = 10 J
Charge (Q) = 5 C
Potential (VA) = 25 volt
Potential (VB) = ?
$$ V = V_B - V_A = \frac{W}{Q} $$
$$ V_B - 25 = \frac{80}{5} $$
$$ V_B - 25 = 16 $$
$$ V_B = 16 + 25 = 41 \text{ volt} $$
Formulas to Remember
$$ V = IR $$
$$ R = \frac{V}{I} $$
$$ I = \frac{V}{R} $$
Where:
V = Voltage (Potential difference) = volt
R = Resistance – Ohm (Ω)
I = Electric current – Ampere
Current (I) = 0.5 A
Resistance (R) = ?
$$ R = \frac{V}{I} $$
Put the values in the formula of resistance:
$$ R = \frac{1.5}{0.5} = 3 \text{ Ohm} $$
Resistance (R) = 1200 Ohm
Current (I) = ?
$$ I = \frac{V}{R} $$
$$ I = \frac{220}{1200} = 0.183 A $$
Electric current (I) = 2 A
Potential difference (V) = ?
$$ V = IR $$
$$ V = 2 \times 15 = 30 \text{ volt} $$
Voltage (V1) = 110 volts
Voltage (V2) = 220 volts
Current (I2) = ?
According to Ohm’s law, V=IR
$$ R = \frac{110}{5} = 22 \text{ Ohm} $$
When the voltage is 220 volt, The current will be:
$$ I_2 = \frac{220}{22} = 10 A $$
Current (I) = 2.5 A
Potential difference (V) = ?
$$ V = IR $$
$$ V = 2.5 \times 6 = 15 \text{ volt} $$
Formulas to Remember
$$ R = \rho \frac{l}{A} $$
Where:
R = Resistance - Ohm
\( \rho \) = Resisitivity - Ohm m
l = length – m
A = area of cross section - m²
Area of cross section (A) = \( 3 \times 10^{-4} \text{ m}^2 \)
Resistivity of copper (\( \rho \)) = \( 1.7 \times 10^{-8} \text{ ohm m} \)
Resistance (R) = ?
$$ R = \rho \frac{l}{A} $$
$$ R = 1.7 \times 10^{-8} \times \frac{0.3}{3 \times 10^{-4}} $$
$$ R = \frac{0.51 \times 10^{-8}}{3 \times 10^{-4}} $$
$$ R = \frac{0.17 \times 10^{-8}}{10^{-4}} $$
$$ R = 0.17 \times 10^{-8} \times 10^4 $$
R = \( 0.17 \times 10^{-4} \text{ Ohm} \)
Length (l) = 1m
Resistance (R) = 0.013 ohm
Resistivity (\( \rho \)) = ?
$$ R = \rho \frac{l}{A} $$
Now put the values in the formula:
$$ 0.013 = \rho \frac{1}{1.20 \times 10^{-6}} $$
$$ \rho = 0.013 \times \frac{1.20 \times 10^{-6}}{1} $$
\( \rho = 1.56 \times 10^{-8} \text{ ohm m} \)
Resistance (R) = 25 ohm
Resistivity (\( \rho \)) = \( 1.84 \times 10^{-6} \text{ ohm m} \)
Area of cross section (A) = ?
$$ R = \rho \frac{l}{A} $$
$$ 25 = 1.84 \times 10^{-6} \times \frac{1}{A} $$
$$ A = 1.84 \times 10^{-6} \times \frac{1}{25} $$
A = \( 7.36 \times 10^{-8} \text{ m}^2 \)
Resistance (R) = 0.85 ohm
Diameter (d) = 0.2 mm = \( 2 \times 10^{-4} \text{m} \)
Radius (r) = d/2 = \( 1 \times 10^{-4} \text{m} \)
Area of cross section (A) = ? (\( \pi r^2 \))
$$ R = \rho \frac{l}{A} $$
$$ R = \rho \frac{l}{\pi r^2} $$
$$ 0.85 = \rho \frac{7}{22 \times (1 \times 10^{-4})^2} \times 1 $$
$$ \rho = \frac{0.85 \times 22}{7 \times 10^{-8}} $$
\( \rho = 2.67 \times 10^{-8} \text{ ohm m} \)
Formulas to Remember
$$ R = R_1 + R_2 + R_3 + \dots R_n $$
$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots \frac{1}{R_n} $$
Where:
\( R_1, R_2, R_3 \dots R_n \) are different resistances
R = total resistance of the combination
R2 = 10 ohm
R3 = 15 ohm
R = ?
$$ R = R_1 + R_2 + R_3 $$
$$ R = 5 + 10 + 15 = 30 \text{ ohm} $$
R2 = 10 ohm
R3 = 15 ohm
R = ?
$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$
$$ \frac{1}{R} = \frac{1}{5} + \frac{1}{10} + \frac{1}{15} $$
$$ \frac{1}{R} = \frac{6+3+2}{30} = \frac{11}{30} $$
$$ \frac{1}{R} = \frac{11}{30} $$
$$ R = \frac{30}{11} = 2.72 \text{ ohm} $$
R2 = 10 ohm
R3 = 15 ohm
V = 90 volt
I = ?
$$ I = \frac{V}{R} $$
We don’t have the value of ‘R’ so first we will find the value:
$$ R = 5 + 10 + 15 = 30 \text{ ohm} $$
$$ I = \frac{V}{R} = \frac{90}{30} = 3 A $$
R2 = 10 ohm
R3 = 15 ohm
V = 100 volt
I1 = ?
I2 = ?
I3 = ?
$$ I_1 = \frac{V}{R_1} $$
$$ I_1 = \frac{100}{5} = 20 A $$
$$ I_2 = \frac{V}{R_2} $$
$$ I_2 = \frac{100}{10} = 10 A $$
$$ I_3 = \frac{V}{R_3} $$
$$ I_3 = \frac{100}{15} = 6.66 A $$
(Note: Current in all resistors is different in the parallel combination.)
R2 = 10 ohm
R3 = 15 ohm
V = 110 volt
I = 10 A
V1 = ?
V2 = ?
V3 = ?
$$ V_1 = I R_1 $$
$$ V_1 = 10 \times 5 = 50 \text{ volt} $$
$$ V_2 = I R_2 $$
$$ V_2 = 10 \times 10 = 100 \text{ volt} $$
$$ V_3 = I R_3 $$
$$ V_3 = 10 \times 15 = 150 \text{ volt} $$
R2 = 3 ohm
R3 = 6 ohm
R = Total resistance = ?
For series combination:
$$ R = R_1 + R_2 + R_3 $$
$$ R = 2 + 3 + 6 = 11 \text{ ohm} $$
For parallel combination:
$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$
$$ \frac{1}{R} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} $$
$$ \frac{1}{R} = \frac{6+4+2}{12} = \frac{12}{12} $$
R = 1 ohm
Current (I) = 0.3 A
Resistance of bulb (r) = 50 ohm
R1 = resistance form variable resistor = ?
Let R1 is the resistance from variable resistor that is used.
Both R1 and r are is series combination so the resultant resistance is:
$$ R = R_1 + r $$
According to Ohm’s law V=IR
$$ 15 = 0.3 \times (R_1 + r) $$
$$ R_1 + 50 = \frac{18}{0.3} = 60 $$
$$ R_1 = 60 - 50 = 10 \text{ ohm} $$
R2 = 10 ohm
R3 = 20 ohm
R = ?
All resistances in the given circuit are connected in the series so total resistance:
$$ R = R_1 + R_2 + R_3 $$
$$ R = 5 + 10 + 20 = 35 \text{ ohm} $$
R2 = 10 ohm
R3 = 20 ohm
R = ?
$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$
$$ \frac{1}{R} = \frac{1}{5} + \frac{1}{10} + \frac{1}{20} $$
$$ \frac{1}{R} = \frac{4+2+1}{20} = \frac{7}{20} $$
$$ R = \frac{20}{7} = 2.85 \text{ ohm} $$
(i) Electric current through each resistor
(ii) Total resistance
(iii) Total current
R2 = 10 ohm
R3 = 30 ohm
V = 6 volt
R = ?
I1 = ?, I2 = ? and I3 = ?
I = ?
All the resistors are connected in parallel combination.
(i) Current through each resistor
$$ V = I_1 R_1 $$
$$ 6 = I_1 \times 5 $$
$$ I_1 = \frac{6}{5} = 1.2 A $$
$$ V = I_2 R_2 $$
$$ 6 = I_2 \times 10 $$
$$ I_2 = \frac{6}{10} = 0.6 A $$
$$ V = I_3 R_3 $$
$$ 6 = I_3 \times 30 $$
$$ I_3 = \frac{6}{30} = 0.2 A $$
(ii) Total resistance (R)
$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$
$$ \frac{1}{R} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30} $$
$$ \frac{1}{R} = \frac{6 + 3 + 1}{30} = \frac{10}{30} $$
$$ R = \frac{30}{10} = 3 \text{ Ohm} $$
(iii) Total current (I)
$$ V = IR $$
$$ 6 = I \times 3 $$
$$ I = \frac{6}{3} = 2 A $$
R2 = 6 ohm
V = 4.5 volt
R = ?
We can see that the resistors of 3 ohm and 6 ohm are connected in parallel combination so the equivalent resistance R
$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} $$
$$ \frac{1}{R} = \frac{1}{3} + \frac{1}{6} $$
$$ \frac{1}{R} = \frac{2 + 1}{6} = \frac{3}{6} $$
$$ R = \frac{6}{3} = 2 \text{ ohm} $$
According to Ohm’s law, V=IR
$$ 4.5 = I \times 2 $$
$$ I = \frac{4.5}{2} = 2.25 A $$
Calculate:
(i) total resistance of the circuit
(ii) total current
(iii) voltage across 5 ohm resistor
R2 = 10 ohm
R3 = 5 ohm
V = 6 volt
(i) total resistance
We can see R1 and R2 are in parallel so the total resistance RA
$$ \frac{1}{R_A} = \frac{1}{R_1} + \frac{1}{R_2} $$
$$ \frac{1}{R_A} = \frac{1}{10} + \frac{1}{10} $$
$$ \frac{1}{R_A} = \frac{1+1}{10} = \frac{2}{10} $$
$$ R_A = \frac{10}{2} = 5 \text{ Ohm} $$
Now RA and R3 in series so the total resistance R
R = RA + R3
R = 5 + 5 = 10 ohm
(ii) total current
$$ I = \frac{V}{R} = \frac{6}{10} = 0.6 A $$
(iii) Voltage across 5 ohm resistor
V1 = IR1
V1 = 0.6 × 5 = 3 volt
R2 = 8 Ohm
R3 = 8 Ohm
R = ?
In the given circuit diagram we can see R2 and R3 are series so:
RA = R2 + R3
RA = 8 + 8 = 16 Ohm
Now the RA is parallel to R1 so the equivalent resistance R
$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_A} $$
$$ \frac{1}{R} = \frac{1}{4} + \frac{1}{16} $$
$$ \frac{1}{R} = \frac{4+1}{16} = \frac{5}{16} $$
$$ R = \frac{16}{5} = 3.2 \text{ Ohm} $$
Formulas to Remember
$$ V = \frac{W}{Q} $$
$$ W = V \times Q $$
$$ W = I^2Rt $$
$$ W = \frac{V^2t}{R} $$
Note = We can put E in place of W
$$ P = \frac{W}{t} $$
$$ P = \frac{V^2}{R} $$
$$ P = I^2R $$
Where:
P = power – Watt
W = work – Joule
V = voltage – Volt
I = current – Ampere
Q = charge = Coulomb
Current (I) = 0.5 A
Power (P) = ?
P = VI
P = 110 × 0.5
P = 55 Watt
Current (I) = 100mA = \( 100 \times 10^{-3} \) A
P = VI
P = \( 5 \times 100 \times 10^{-3} \)
P = 0.5 Watt
(ii) P2 = 110 W , 220 volt
R1 = ?
R2 = ?
$$ R_1 = \frac{V^2}{P_1} $$
$$ R_1 = \frac{(220)^2}{60} $$
$$ R_1 = 806.67 \text{ Ohm} $$
$$ R_2 = \frac{V^2}{P_2} $$
$$ R_2 = \frac{(220)^2}{100} $$
$$ R_2 = 484 \text{ Ohm} $$
60 watt bulb has higher resistance than 100 watt bulb.
P2 = 60 W
V = 220volt
I = ?
$$ I = \frac{V}{R} $$
In the above formula we don’t have value of R so we will have to find the value of R
$$ R_1 = \frac{V^2}{P_1} = \frac{(220)^2}{100} \text{ Ohm} $$
$$ R_2 = \frac{V^2}{P_2} = \frac{(220)^2}{60} \text{ Ohm} $$
Both the lamps are in parallel, so the resultant resistance is
$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} $$
$$ \frac{1}{R} = \frac{100}{220 \times 220} + \frac{60}{220 \times 220} $$
$$ \frac{1}{R} = \frac{100+60}{220 \times 220} $$
$$ \frac{1}{R} = \frac{160}{220 \times 220} \text{ Ohm} $$
Now we can find the value of current (I)
$$ I = \frac{220}{\frac{220 \times 220}{160}} $$
$$ I = \frac{220 \times 160}{220 \times 220} $$
$$ I = \frac{160}{220} = 0.72 A $$
V = 220 Volt
$$ \text{Energy consumed per day} = \frac{P \times t}{1000} $$
$$ = \frac{100 \times 10}{1000} $$
Energy consumed per day = 1KWh or 1 Unit
I = 650 mA = 0.65A
T = 5 hours
(i) P=VI
P = 2.5 × 0.65 = 1.625 W
(ii) Resistance
$$ R = \frac{V}{I} $$
$$ R = \frac{2.5}{0.65} = 3.84 \text{ Ohm} $$
(iii) Energy consumed in 5 hours
$$ = \frac{P \times t}{1000} $$
$$ = \frac{1.625 \times 5}{1000} = 0.008125 \text{ KWh} $$
T = 1 hour
(i) Energy consumed in KWh
$$ \text{Energy consumed for 1 hour} = \frac{P \times t}{1000} $$
$$ = \frac{750 \times 1}{1000} = 0.75 \text{ KWh} $$
(iii) Energy consumed in Joule
1 KWh = \( 3.6 \times 10^6 \) Joule
So \( 0.75 \times 3.6 \times 10^6 \) Joule
= \( 2.7 \times 10^6 \) Joule
I = 5 A
$$ \text{Energy consumed for 5 hour} = \frac{P \times t}{1000} $$
But in this formula we need value of Power (P)
P = VI
P = 220 × 5
P = 1100W
Now we can put the value of P in the formula
$$ \text{Energy consumed for 5 hour} = \frac{1100 \times 5}{1000} $$
= 5.5 KWh
T1 = 1hour
P2 = 1200W
T2 = 10 min = 10/60 hour
(i) A 250 W computer is used for 1 hour
= P1 × t1 = 250 × 1
= 250Wh
(ii) A 1200W heater is used for 10 min
= P2 × t2 = 1200 × 10/60
= 200Wh
So, computer uses more energy.
V = 220 volt
R = ?
$$ R = \frac{V^2}{P} $$
$$ R = \frac{220^2}{1000} $$
R = 48.4 Ohm
Voltage (V) = 220volt
Resistance (R) = ?
$$ P = \frac{V^2}{R} $$
$$ 100 = \frac{220^2}{R} $$
$$ R = \frac{220^2}{100} $$
R = 484 Ohm
V = 15 volt
T = 1 min = 60 s
H = P × t
H = 15 × 60 = 900 J
Time (t) = 1 s
Resistance (R) = 50 Ohm
Voltage (V) = ?
$$ H = I^2Rt $$
$$ 1000 = I^2 \times 50 \times 1 $$
$$ I^2 = \frac{1000}{50} = 20 $$
$$ I = \sqrt{20} = 4.47A $$
So, potential difference across the resistor
V = IR
V = 4.47 × 50 = 223.5 volt
Current (I) = 10 A
Time (t) = 2 hours = 2 × 60 × 60 = 7200 s
Heat (H) = ?
$$ H = I^2Rt $$
$$ H = (10)^2 \times 10 \times 7200 $$
H = \( 72 \times 10^5 \) J
Time (t) = 1s
Resistance (R) = 10 Ohm
Potential difference (V) = ?
V = IR
In the above formula we need the value of current (I) so at first we must find the value of I form the given values
$$ H = I^2Rt $$
$$ I = \sqrt{\frac{H}{Rt}} $$
$$ I = \sqrt{\frac{100}{10 \times 1}} $$
$$ I = \sqrt{10} $$
$$ I = 3.16 A $$
Now we can find the value of potential difference
V = IR
V = 3.16 × 10
V = 31.6 volt
R2 = 2 Ohm
Potential difference (V) = 10 volt
Time (t) = 10 s
Heat (H) = ?
H = V I t
We will have to find the value of current (I)
R = R1 + R2
R = 2 + 2 = 4 Ohm
$$ I = \frac{V}{R} $$
$$ I = \frac{10}{4} = 2.5 A $$
H = V I t
H = 10 × 2.5 × 10
H = 250 J
Voltage (V) = 15 volt
Energy (E) = ?
E = VQ
E = 15 × 10 = 150 J