If a bulb of 60 W is connected across a source a source of 220 V , find the current drawn by it.

Solution : Data : P=60 W, V=220 V, I=?

P=VI

∴ I=P/V=60W/220V=(3/11)A = 0.2727A

The current drawn by the bulb = 0.2727A (Approx)

If two resistors are connected in series, the total resistance is 45 Ω and if the same resistors are connected in parallel, the total resistance becomes 10 Ω . Find the values of individual resistors. Solution: Data : Rs = 45Ω , Rp = 10Ω , R1 = ? , R2 = ?

In Series combination: R_s = R₁ + R₂ = 45Ω  ........ eq. no. (1)

∴ R₂ = 45 – R₁ ......... eq. no. (2)

In Parallel combination : 1/R_p = 1/R₁ + 1/R₂ = 1/10Ω

∴ (R₁ + R₂)/R₁R₂ = 1/10Ω

∴ R₁R₂/(R₁ + R₂) = 10Ω

∴ R₁ (45 – R₁) / 45 = 10   [From eq. no. (1) & (2) ]

∴ 45R₁ – R₁² = 450

∴ R₁² – 45R₁  + 450 = 0

∴ R₁² – 30R₁ – 15R₁ + 450 = 0

∴ R₁ (R₁ – 30) – 15 (R₁ – 30) = 0

∴ (R₁ – 30) (R₁ – 15) = 0

∴ R₁ – 30 = 0    OR   R₁ – 15 = 0

∴ R₁ = 30   OR   R₁ = 15

∴ The values of individual resistors are 30Ω and 15Ω

Find the effective resistance of the combination of two resistors of resistances 10 Ω and 15 Ω connected in (i) Series (ii) Parallel.

Solution: Data: R₁ = 10 Ω , R₂ = 15Ω , R_s = ?, R_p = ?

(i) R_s = R₁ + R₂ = 10 Ω + 15Ω = 25Ω

The effective resistance in the series combination = 25 Ω

(ii) 1/R_p = 1/R₁ + 1/R₂ = 1/10Ω +1/15Ω = 5/30Ω

∴ R_p = 30Ω/5 = 6Ω

The effective resistance in the parallel combination = 6 Ω. 

The resistance of a wire of length 31.4 m and diameter 1 mm is 20 Ω . Find the resistivity of the material of the wire.

Solution: Data : l = 31.4 m, d = 1mm = 1× 10 ^{- 3} m

∴ r = d/2 = (1× 10^-3m)/2 = 5× 10^-4 m , R = 20 Ω , ρ = ?

R = ρ (l/A) = ρ (l/π r²)

∴ ρ = (Rπr²)/l

∴ ρ = 20Ω × 3.14 × (5× 10^-4 m)² / 31.4m

∴ ρ = 5× 10^-7Ω .m

The resistivity of the material of the wire = 5 × 10^-7 Ω m.

Calculate the potential difference across a 10 Ω resistor carrying a current of 0.2 A.

Solution: Data : R =10 Ω , I = 0.2 A, V = ?

V = IR  = 0.2 A × 10 Ω = 2 V

The potential difference across the resistor = 2 V. 

A wire carries a current of 0.2 A for 10 seconds. If the potential difference between the two ends of the wire is 20 V, find the work done in this process.

Solution: Data : I = 0.2 A, t = 10 s, V = 20 V, W = ?

I = Q/t and V = W/Q

∴ W = VQ = VIt = 20 V  × 0.2 A × 10 s = 40 J

The work done = 40 J

If 15J of work is done when a charge of 3C is moved from one point to another, find the electric potential difference between the two points.

Solution: Data: W = 15J, Q = 3 C, V = ?

V = W/Q = 15/3 = 5V

The electric potential difference between the two points = 5V. 

An electric charge of 15C passes through the cross section of a wire in one minute. Find the current through the wire.

Solution: Data: Q = 15C, t = 1 minute = 60 s, I = ?

I = Q/t = 15/60 = 0.25A

∴ The current through the wire = 0.25 A.