Question

NUMERICALS
1. An electric bulb of resistance 500 2 draws current
0-4 A from the source. Calculate : (a) the power of

bulb and (b) the potential difference at its end.​

Answer 1

Answer:

• Power of bulb is 80 Watt
• Potential Difference is 200 volts.

Explanation:

Given:-

• Resistance ,R = 500 Ω

• Current ,I = 0.4 A

To Find:-

• Power ,P
• Potential Difference ,V

Solution:-

(b)

Firstly we calculate the potential difference

Using Ohm's Law

• V = IR

•❒ V = IR

Where ,

• V denotes Voltage or Potential Difference

• I denotes current

• R denotes Resistance

Substitute the value we get

→ V = 0.4×500

→ V = 200V

• Hence, potential difference between at its end is 200 volts.

(a)

Now, calculating the power of the bulb.

As we know that

• P = VI

Where,

• P denotes Power
• V denote voltage
• I denotes Current

Substitute the value we get

→ P = 200 × 0.4

→ P = 20×4

→ P = 80 W

• Hence, the power of the bulb is 80 watt.

Answer 2

Answer:

Given :-

• Resistance = 500 Ω
• Current = 0.4 A

To Find :-

• Power
• Potential difference

Solution :-

We will use Ohm's law for finding potential difference

${\huge {\fbox {\pink{V = IR}}}}$

Here,

V indicates Potential difference

I indicates Current

R indicate Resistance

$\tt \: V = IR$

$\tt \: V = 500 \times 0.4$

$\tt \: V = 200 V$

Now,

Let's find power

${\huge {\fbox {\pink {P = VI}}}}$

Here,

P indicate Power

V indicates Potential difference

I indicates Current

$\tt \: P = 200 \times 0.4$

$\tt \: P = 80 W$

Hence :-

• Potential difference of bulb is 200 Volt
• Power of bulb is 80 W