A cell of emf 90 V is connected across series combination of two resistors each of 100 Ω resistance. A voltmeter of resistance 400 Ω is used to measure the potential difference across one of the given resistors. The reading of the voltmeter will be:
Question
A cell of emf 90 V is connected across series combination of two resistors each of 100 Ω resistance. A voltmeter of resistance 400 Ω is used to measure the potential difference across one of the given resistors. The reading of the voltmeter will be:
Solution
Step 1: Identify the given values
The emf of the cell, E = 90 V The resistance of each resistor in the series, R = 100 Ω The resistance of the voltmeter, V = 400 Ω
Step 2: Calculate the total resistance in the circuit
Since the two resistors are in series, their resistances add up. So, the total resistance in the circuit, R_total = R1 + R2 = 100 Ω + 100 Ω = 200 Ω
Step 3: Calculate the current in the circuit
Using Ohm's law, I = E / R_total = 90 V / 200 Ω = 0.45 A
Step 4: Calculate the potential difference across one resistor
The potential difference across one resistor, V_R = I * R = 0.45 A * 100 Ω = 45 V
Step 5: Calculate the potential difference measured by the voltmeter
The voltmeter is connected in parallel to one of the resistors. Therefore, the potential difference it measures is the same as the potential difference across the resistor it is connected to. So, the reading of the voltmeter, V_voltmeter = V_R = 45 V
So, the reading of the voltmeter will be 45 V.
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