Question

The voltage of a bulb is 24w. When it is connected to 12v battery. Calculate it's effective. Voltage if it operates on a 6v battery.

Answer 1

Correct Question

The voltage of a bulb is 24W. When it is connected to 12V battery. Calculate it's effective resistance and power if it operates on a 6V battery.

Solution

We know that, P = V²/R

Given that if V is 12, then P is 24.

So, effective resistance R would be

P = V²/R

→ R = V²/P

Substitute the values

→ R = (12)²/24

→ R = 144/24

→ R = 6 Ω

So, effective resistance of the bulb is 6 Ω.

Now, given to find power when V is 6.

We have derived R as 6

So, P = V²/R

→ P = 6²/6

→ P = 36/6

→ P = 6 W

Hence, the power generated in bulb by 6 V battery is 6 W.

Answer 2

» The voltage of current is 24 W, when connected to 12 V battery.

Power (P) = 24 V

Voltage (V) = 12 V

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

R = $\dfrac{ {12}^{2} }{P}$

R = $\dfrac{144}{24}$

R = 6 ohm

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» Calculate it's effective resistance (R) when it operates on 6V battery.

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

We have ...

V = 6V

R = 6 ohm (from above calculations)

P = $\dfrac{ {6}^{2} }{6}$

P = $\dfrac{36}{6}$

P = 6 W

_____________ [ANSWER]

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✡ Derivation of used formula :

» P = V × I ____ (eq 1)

Here...

• P = Potential difference

• V = Voltage

• I = current

» V = $\frac{I}{R}$ [According to Ohm's Law]

I = VR

Put value of I in (eq 1)

=> P = V × $\frac{V}{R}$

=> P = $\dfrac{ {V}^{2} }{R}$

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