Q165 / 180 · KCET 2023 · Physics

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
    $2.4 \mathrm{~A}$
  • Correct answer
    C
    $0.6 \mathrm{~A}$.
  • D
    $1.2 \mathrm{~A}$

Solution

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, $I=\frac{E_{\mathrm{net}}}{R^{\prime}}=\frac{12}{20}=0.6 \mathrm{~A}$

Hence, the correct answer is option: 'C'.

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