KCET · Physics · Current Electricity
Ten identical cells each emf \(2 \mathrm{~V}\) and internal resistance \(1 \Omega\) are connected in series with two cells wrongly connected. A resistor of \(10 \Omega\) is connected to the combination. What is the current through the resistor?
- A \(1.8 \mathrm{~A}\)
- B \(2.4 \mathrm{~A}\)
- C \(0.6 \mathrm{~A}\).
- D \(1.2 \mathrm{~A}\)
Answer & Solution
Correct Answer
(C) \(0.6 \mathrm{~A}\).
Step-by-step Solution
Detailed explanation
According to given situation,
Net emf of the combination,
\(E_{\text {net }}=8 \times\) emf of one cell \(-2 \times\) emf of one cell
\(=8 \times 2-2 \times 2=12 \mathrm{~V}\)
Equivalent resistance of the circuit
\(R^{\prime}=10 r+R=10 \times 1+10=20 \mathrm{~V}\)
\(\therefore\) Current through the resistor,
\(I=\frac{E_{\mathrm{net}}}{R^{\prime}}=\frac{12}{20}=0.6 \mathrm{~A}\)
Net emf of the combination,
\(E_{\text {net }}=8 \times\) emf of one cell \(-2 \times\) emf of one cell
\(=8 \times 2-2 \times 2=12 \mathrm{~V}\)
Equivalent resistance of the circuit
\(R^{\prime}=10 r+R=10 \times 1+10=20 \mathrm{~V}\)
\(\therefore\) Current through the resistor,
\(I=\frac{E_{\mathrm{net}}}{R^{\prime}}=\frac{12}{20}=0.6 \mathrm{~A}\)
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