Solved Problems on Electricity: Calculating Resistance, Current, and Power

Solved Problems - Electricity | Science

Electricity - Science

Solved Problems

1. Two bulbs are having the ratings as 60 W, 220 V and 40 W, 220 V respectively. Which one has a greater resistance?

Solution:

Electric power P is given by the formula:

$$P = \frac{V^2}{R}$$

For the same value of V, R is inversely proportional to P.

Therefore, lesser the power, greater the resistance.

Hence, the bulb with 40 W, 220 V rating has a greater resistance.

2. Calculate the current and the resistance of a 100 W, 200 V electric bulb in an electric circuit.

Solution:

Power P = 100 W and Voltage V = 200 V

Power P = V $\times$ I

3. In the circuit diagram given below, three resistors $R_1$, $R_2$ and $R_3$ of 5 Ω, 10 Ω and 20 Ω respectively are connected as shown. Calculate:

1. Current through each resistor
2. Total current in the circuit
3. Total resistance in the circuit

Solution:

A) Since the resistors are connected in parallel, the potential difference across each resistor is same (i.e. V=10V).

Therefore, the current through $R_1$ is,

B) Total current in the circuit, $I = I_1 + I_2 + I_3$

= 2 + 1 + 0.5 = 3.5 A

4. Three resistors of 1 Ω, 2 Ω and 4 Ω are connected in parallel in a circuit. If a 1 Ω resistor draws a current of 1 A, find the current through the other two resistors.

Solution:

$R_1$ = 1 Ω, $R_2$ = 2 Ω, $R_3$ = 4 Ω. Current $I_1$ = 1 A

The potential difference across the 1 Ω resistor = $I_1 R_1 = 1 \times 1 = 1 V$

Since, the resistors are connected in parallel in the circuit, the same potential difference will exist across the other resistors also.

So, the current in the 2 Ω resistor, $\frac{V}{R_2} = \frac{1}{2} = 0.5 A$

Similarly, the current in the 4 Ω resistor, $\frac{V}{R_3} = \frac{1}{4} = 0.25 A$

Electric Power: Definition, Formula, SI Unit, and Energy Consumption Explained

Electric Power - Definition, Formula, Unit, Consumption

ELECTRIC POWER

In general, power is defined as the rate of doing work or rate of spending energy. Similarly, the electric power is defined as the rate of consumption of electrical energy. It represents the rate at which the electrical energy is converted into some other form of energy.

Suppose a current ‘I’ flows through a conductor of resistance ‘R’ for a time ‘t’, then the potential difference across the two ends of the conductor is ‘V’. The work done ‘W’ to move the charge across the ends of the conductor is given by the equation (4.19) as follows:

$$ W = VIt $$

$$ P = \frac{\text{Work}}{\text{Time}} = \frac{VIt}{t} $$

$$ P = V I \quad (4.21) $$

Thus, the electric power is the product of the electric current and the potential difference due to which the current passes in a circuit.

1. Unit of Electric Power

The SI unit of electric power is watt. When a current of 1 ampere passes across the ends of a conductor, which is at a potential difference of 1 volt, then the electric power is

\( P = 1 \text{ volt} \times 1 \text{ ampere} = 1 \text{ watt} \)

Thus, one watt is the power consumed when an electric device is operated at a potential difference of one volt and it carries a current of one ampere. A larger unit of power, which is more commonly used is kilowatt.

2. Consumption of electrical energy

Electricity is consumed both in houses and industries. Consumption of electricity is based on two factors: (i) Amount of electric power and (ii) Duration of usage. Electrical energy consumed is taken as the product of electric power and time of usage. For example, if 100 watt of electric power is consumed for two hours, then the power consumed is 100 × 2 = 200 watt hour. Consumption of electrical energy is measured and expressed in watt hour, though its SI unit is watt second. In practice, a larger unit of electrical energy is needed. This larger unit is kilowatt hour (kWh) . One kilowatt hour is otherwise known as one unit of electrical energy. One kilowatt hour means that an electric power of 1000 watt has been utilized for an hour. Hence,

$$ 1 \text{ kWh} = 1000 \text{ watt hour} = 1000 \times (60 \times 60) \text{ watt second} = 3.6 \times 10^6 \text{ J} $$