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 determine the voltmeter reading across the 200 \((\Omega)\) resistor, the total resistance of the series circuit must first be computed. The total resistance \(R_{\text{total}}\) is the sum of the individual resistances:

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

Ohm's Law is then applied to calculate the total current \(I\) flowing through the circuit:

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

The voltage drop across the 200 \((\Omega)\) resistor is subsequently calculated using Ohm's Law:

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

Consequently, the voltmeter reading is \(\frac{40}{3}\,V\).

The correct answer is therefore \(\frac{40}{3}\,V\).