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. \(\frac{20}{3}V\)
• C. 10V
• D. 16V

Answer 1

The Correct Option is B

Given: Two resistances of 100 Ω and 200 Ω are connected in series across a 20 V battery. The voltage reading across the 200 Ω resistance is to be determined.

Step 1: Understanding the Circuit

The resistors \( R_1 = 100 \, \Omega \) and \( R_2 = 200 \, \Omega \) are connected in series with a 20 V battery.

In a series circuit, the current is the same through all components. Using Ohm's Law, the total resistance \( R_{\text{total}} \) in the series circuit is:

Rtotal = R1 + R2

Rtotal = 100 Ω + 200 Ω = 300 Ω

Step 2: Calculating the Current

Using Ohm’s law, the total current \( I \) in the circuit can be calculated as:

I = V / Rtotal

Substituting the values:

I = 20 V / 300 Ω = 1/15 A

The current flowing through the circuit is \( \frac{1}{15} \, \text{A} \).

Step 3: Voltage Across the 200 Ω Resistor

Now, to find the voltage across the 200 Ω resistor, we use Ohm’s Law again:

V = I * R2

Substituting the values:

V = (1/15) A * 200 Ω = 200/15 V = 20/3 V

Step 4: Final Answer

The voltage across the 200 Ω resistor is \( \frac{20}{3} \, \text{V} \), which is approximately 6.67 V.

Correct Option: Option 2: 20/3 V