Question 181: Current Electricity

Question

The voltage and the current between $A$ and $B$ points shown in the circuit are $\_\_\_\_$ .

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

$$ 24 \mathrm{~V}, 12 \mathrm{~A} $$

Answer

$$ 24 \mathrm{~V}, 4 \mathrm{~A} $$

Answer

$$
18 \mathrm{~V}, 12 \mathrm{~A}
$$

Answer

$$ 27 \mathrm{~V}, 4 \mathrm{~A} $$

Current Electricity

The voltage and the current between A and B points shown in the circuit are \_\_\_\_ .

Choose the correct answer: