Question

In the circuits shown below, the readings of the voltmeters and the ammeters will be:

• A. \(V_2>V_1\) and \(i_1=i_2\)
• B. \(V_1=V_2\) and \(i_1>i_2\)
• C. \(V_1=V_2\) and \(i_1=i_2\)
• D. \(V_2>V_1\) and \(i_1=i_2\)

Answer 1

The Correct Option is C

To determine the readings of the voltmeters (V_1 and V_2) and ammeters (i_1 and i_2) in the given circuits, let's analyze each circuit step-by-step.

Analysis of Circuit 1:

• Circuit 1 consists of a single resistor of 10 \, \Omega connected in series with a 10 V battery.
• According to Ohm's Law, V = I \times R. Thus, the current i_1 through the circuit can be calculated as follows:
• i_1 = \frac{V}{R} = \frac{10 \, \text{V}}{10 \, \Omega} = 1 \, \text{A}
• The voltmeter V_1 across the resistor will read the total battery voltage of 10 V as it measures the potential difference across the resistor, which is the same as the battery voltage in this simple series circuit.

Analysis of Circuit 2:

• Circuit 2 has two 10 \, \Omega resistors in parallel, connected to a 10 V battery.
• The total resistance R_t in parallel is calculated as:
• \frac{1}{R_t} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10}, hence R_t = 5 \, \Omega
• The total current i_2 flowing from the battery is:
• i_2 = \frac{V}{R_t} = \frac{10 \, \text{V}}{5 \, \Omega} = 2 \, \text{A}
• This total current splits equally between the two parallel resistors. Therefore, each branch carries 1 \, \text{A}.
• The voltmeter V_2, like V_1, measures across one resistor in a parallel setup, showing 10 \, \text{V}.

Conclusion:

Both voltmeters read the full voltage of 10 \, \text{V} and both ammeters read 1 \, \text{A} as the current through each branch. Therefore:

V_1 = V_2 and i_1 = i_2