Question 116: Current Electricity

Question

Two resistors of $100 \Omega$ each are connected in series with a 9 V battery. A voltmeter of $400 \Omega$ resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is $\_\_\_\_$ V.

Accepted Answer

The initial circuit is,
When the voltmeter is connected to measure the voltage across one resistor, it is connected in parallel with that resistor.
Resistance of Voltmeter $\mathrm{R}_{\mathrm{V}}=400 \Omega$
The voltmeter $\left(\mathrm{R}_{\mathrm{V}}ight)$ and the resistor $\left(\mathrm{R}_1ight)$ are in parallel.
$$ \frac{1}{R_{1, V}}=\frac{1}{R_1}+\frac{1}{R_V} $$
$$\Rightarrow $$ $\frac{1}{R_{1, V}}=\frac{1}{100}+\frac{1}{400}=\frac{5}{400}$
$$ \mathrm{R}_{1, \mathrm{~V}}=\frac{400}{5}=80 \Omega $$
Now, $R_2$ is in series with the equivalent resistance $\left(R_{1, V}ight)$. So, the total resistance of the circuit is,
$$ \mathrm{R}_{	ext {total }}=\mathrm{R}_{1, \mathrm{~V}}+\mathrm{R}_2 $$
$$\Rightarrow $$ $$ \mathrm{R}_{	ext {total }}=100 \Omega+80 \Omega=180 \Omega $$
Using Ohm's Law ( $\mathrm{V}=\mathrm{I} \cdot \mathrm{R}$ ) for the entire circuit :
$$ \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{	ext {total }}} $$
$$\Rightarrow $$ $$ I=\frac{9}{180}=\frac{1}{20}=0.05 \mathrm{~A} $$
The voltmeter measures the potential difference across the parallel section ( $\mathrm{R}_{1, \mathrm{~V}}$ ).
Using Ohm's Law for that specific section :
$$ V_{	ext {reading }}=I 	imes R_{1, V} $$
$$\Rightarrow $$ $\mathrm{V}_{	ext {reading }}=0.05 \mathrm{~A} 	imes 80 \Omega$
$$\Rightarrow $$ $$ \mathrm{V}_{	ext {reading }}=4 \mathrm{~V} $$
Therefore, the voltmeter reading is 4 V. Hence, the correct option is (C).

Answer

3

Answer

2

Answer

4

Answer

4.5

Current Electricity

Two resistors of 100 \Omega each are connected in series with a 9 V battery. A voltmeter of 400 \Omega resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is \_\_\_\_ V.

Choose the correct answer: