A $8 \Omega$ resistor is connected to a battery that has an internal resistance of $0.2 \Omega$. If the…

A $8 \Omega$ resistor is connected to a battery that has an internal resistance of $0.2 \Omega$. If the voltage across the battery (the terminal voltage) is $10 \mathrm{~V}$, then the emf of the battery is
  1. $10.15 \mathrm{~V}$
  2. $10.20 \mathrm{~V}$
  3. $10.25 \mathrm{~V}$
  4. $9.80 \mathrm{~V}$

Solution

The given situation is shown below
If $I=$ current in circuit then Where, $ \begin{gathered} E-I r=V \\ I=V / R \end{gathered} $ Here, given, $ \begin{aligned} & R=8 \Omega \\ & r=0.2 \Omega V=10 \mathrm{~V} \end{aligned} $ Hence, circuit current $ I=\frac{V}{R}=\frac{10}{8}=\frac{5}{4} \mathrm{~A} $ Now, from eq. (i) with values we have, $ E=V+I r=10+\frac{5}{4} \times 0.2=10+0.25=10.25 \mathrm{~V} $

Asked in: AP EAMCET 2022 (07 Jul Shift 2)

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