Question

Two resistances of 100 \(\Omega\) and 200 \(\Omega\) are connected in series across a 20 V battery as shown in the figure below. The reading in a 200 \(\Omega\) voltmeter connected across the 200 \(\Omega\) esistance is _______.

Fill in the blank with the correct answer from the options given below

• A. 4V
• B. \(\frac{40}{3}V\)
• C. 10V
• D. 16V

Answer 1

The Correct Option is B

To find the reading on the voltmeter connected across the 200 \((\Omega)\) resistor, we must first calculate the total resistance in the series circuit. The total resistance \(R_{\text{total}}\) is given by the sum of the individual resistances: 

\(R_{\text{total}} = 100\,\Omega + 200\,\Omega = 300\,\Omega\)

Using Ohm's Law, the total current \(I\) flowing through the circuit is calculated as follows:

\(I = \frac{V_{\text{battery}}}{R_{\text{total}}} = \frac{20\,V}{300\,\Omega} = \frac{1}{15}\,A\)

Now, calculate the voltage drop across the 200 \((\Omega)\) resistor using Ohm's Law:

\(V_{200} = I \times 200\,\Omega = \frac{1}{15}\,A \times 200\,\Omega = \frac{200}{15}\,V = \frac{40}{3}\,V\)

The voltmeter reading is therefore \(\frac{40}{3}\,V\).

Thus, the correct answer is \(\frac{40}{3}\,V\).