AP EAMCET · PHYSICS · Current Electricity
Two resistances of \(400 \Omega\) and \(800 \Omega\) are connected in series with \(6 \mathrm{~V}\) battery of negligible internal resistance. A voltmeter of resistance \(10000 \Omega\) is used to measure the potential difference across \(400 \Omega\). The error in the measurement of potential difference in volts approximately is :
- A (a) 0.01
- B 0.02
- C 0.03
- D 0.05
Answer & Solution
Correct Answer
(D) 0.05
Step-by-step Solution
Detailed explanation
\(R_1=400 \Omega, R_2=800 \Omega\) \(P D\) across \(400 \Omega\) resistance (when voltmeter is not connected) \(V_1=\frac{6}{(400+800)} \times 400\) \(=\frac{6 \times 400}{1200}=2 \mathrm{~V}\) when voltmeter is connected Total resistance of the circuit,…
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