Question 182: Current Electricity

Question

Two resistors of 200 $\Omega$ and 400 $\Omega$ are connected in series with a battery of 100 V. A bulb rated at 200 V, 100 W is connected across the 400 $\Omega$ resistance. The potential drop across the bulb is ________ V.

Accepted Answer

First, find the resistance of the bulb using its rating.
Given for the bulb:
Voltage, $V = 200\ 	ext{V}$
Power, $P = 100\ 	ext{W}$
We know,
$$ R = \frac{V^2}{P} $$
So,
$$ R_b = \frac{200^2}{100} = \frac{40000}{100} = 400\ \Omega $$
So the bulb has resistance $400\ \Omega$.
Now the bulb is connected across the $400\ \Omega$ resistor, so these two are in parallel.
Hence we have:
one resistor of $200\ \Omega$ in series with
parallel combination of $400\ \Omega$ and $400\ \Omega$
Now calculate the equivalent resistance of the parallel part:
$$ R_p = \frac{400 	imes 400}{400 + 400} = \frac{160000}{800} = 200\ \Omega $$
So total resistance of the circuit is:
$$ R_{	ext{total}} = 200 + 200 = 400\ \Omega $$
Now total current from the battery:
$$ I = \frac{V}{R_{	ext{total}}} = \frac{100}{400} = 0.25\ 	ext{A} $$
The voltage drop across the series $200\ \Omega$ resistor is:
$$ V_1 = IR = 0.25 	imes 200 = 50\ 	ext{V} $$
Therefore, the remaining voltage across the parallel combination is:
$$ V_p = 100 - 50 = 50\ 	ext{V} $$
Since the bulb is connected in parallel with the $400\ \Omega$ resistor, the potential difference across the bulb is also:
$$ \boxed{50\ 	ext{V}} $$
So, the correct option is
B
.

Answer

25

Answer

50

Answer

66.6

Answer

100

Current Electricity

Two resistors of 200 \Omega and 400 \Omega are connected in series with a battery of 100 V. A bulb rated at 200 V, 100 W is connected across the 400 \Omega resistance. The potential drop across the bulb is ________ V.

Choose the correct answer: