Numericals Physics Physics problems

Combination of resistors

May 15, 2024
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To solve this problem, we need to determine the reading of a voltmeter with a resistance of 400Ω when connected across a 100Ω resistor, which is in series with a 200Ω resistor. Both resistors are connected to an 84V source. We’ll also calculate the potential difference across the 100Ω resistor before the voltmeter was connected.

### Part (a): Voltmeter Reading

When the voltmeter is connected across the 100Ω resistor, the voltmeter and the resistor form a parallel combination. This parallel branch is in series with the 200Ω resistor.

1. **Total Resistance of the Parallel Branch**:
– R_{\text{parallel}} = \left(\frac{1}{100} + \frac{1}{400}\right)^{-1}

2. **Total Resistance of the Circuit**:
– R_{\text{total}} = R_{\text{parallel}} + 200

3. **Total Current through the Circuit**:
– I_{\text{total}} = \frac{V_{\text{source}}}{R_{\text{total}}}

4. **Voltage across the Parallel Branch (Voltage read by the Voltmeter)**:
– V_{\text{voltmeter}} = I_{\text{total}} \times R_{\text{parallel}}

### Part (b): Voltage across the 100Ω Resistor Before Connecting the Voltmeter

Without the voltmeter, the resistors 100Ω and 200Ω are in series:
1. **Total Resistance without Voltmeter**:
– R_{\text{total}}' = 100 + 200

2. **Total Current through the Circuit without Voltmeter**:
– I_{\text{total}}' = \frac{V_{\text{source}}}{R_{\text{total}}'}

3. **Voltage across the 100Ω Resistor**:
– V_{100} = I_{\text{total}}' \times 100

Let’s perform the calculations now.

It seems there was an issue with the calculation. Let me try solving it again.

### Part (a): Voltmeter Reading

Given:
– Voltage source V = 84V
– Resistances: R1 = 100\Omega, R2 = 200\Omega, R_{voltmeter} = 400\Omega

For the parallel combination of the 100Ω resistor and the voltmeter:

    \[ R_{parallel} = \left(\frac{1}{100} + \frac{1}{400}\right)^{-1} \]

Total circuit resistance when the voltmeter is connected:

    \[ R_{total} = R_{parallel} + 200 \]

Total current through the circuit:

    \[ I_{total} = \frac{V_{source}}{R_{total}} \]

Voltage across the parallel branch (and the voltmeter reading):

    \[ V_{voltmeter} = I_{total} \times R_{parallel} \]

### Part (b): Voltage across the 100Ω Resistor Before Connecting the Voltmeter

Total resistance of the 100Ω and 200Ω resistors in series:

    \[ R_{total}' = 100 + 200 \]

Total current without the voltmeter:

    \[ I_{total}' = \frac{V_{source}}{R_{total}'} \]

Voltage across the 100Ω resistor before the voltmeter:

    \[ V_{100} = I_{total}' \times 100 \]

Let’s perform these calculations manually, given the system issue:

#### For Part (a):
1. R_{parallel} = \left(\frac{1}{100} + \frac{1}{400}\right)^{-1} = \left(0.01 + 0.0025\right)^{-1} = \left(0.0125\right)^{-1} = 80\Omega
2. R_{total} = 80\Omega + 200\Omega = 280\Omega
3. I_{total} = \frac{84V}{280\Omega} = 0.3A
4. V_{voltmeter} = 0.3A \times 80\Omega = 24V

#### For Part (b):
1. R_{total}' = 300\Omega
2. I_{total}' = \frac{84V}{300\Omega} = 0.28A
3. V_{100} = 0.28A \times 100\Omega = 28V

Based on these manual calculations:
– The reading of the voltmeter (Part a) would be 24V.
– The voltage across the 100Ω resistor before connecting the voltmeter (Part b) would be 28V.