Question

11. 2 A current flows through an electrical device of resistance 120 2. a) Calculate power of the device? (2) b) What changes to be done in the device to increase the power of the device without changing potential difference? (1) 12​

Answer 1

solution

a) Current (I) = 2A ( given )

Resistance (R) = 120 ohm

To calculate = power (p)

Power = current² × resistance

so p = I²R

Therefore , p= 2² ×120 watt

p = 4×120 watt

p = 480 watt

Answer 2

Given,

• Current Flow (I) = 2A
• Resistance (R) = 120 Ω

(a)

Here, we have to find the power of the device.

In terms of I and R,

$\sf\bold{Power (P) = I^2R}$

Applying the values to the equation,

$\sf{\implies P=2^2 \times 120}$

$\sf{\implies P=4 \times 120}$

$\sf{\implies P=480 \ W }$

$\huge{\boxed{\bold{\sf{\therefore \ Power \ of \ the \ device \ = 480 \ W }}}}$

(b)

In order to increase the power of the device without changing potential difference, we have two choices:

• Increase the current flowing through the device.
• Increase the resistance of the device.