Question

Two batteries with e.m.f. $12V$ and $13V$ are connected in parallel across a load resistor of $10\Omega$. The internal resistances of the two batteries are $1\Omega$ and $2\Omega$ respectively. The voltage across the load lies between.

• A. $11.4V$ and $11.5V$
• B. $11.7V$ and $11.8V$
• C. $11.6V$ and $11.7V$
• D. $11.5V$ and $11.6V$

Answer 1

The correct option is D

$11.5V$ and $11.6V$

Step 1: Given Data

Emf of the first battery $E_1=12V$

Emf of the second battery $E_2=13V$

The internal resistance of the first battery $r_1=1\Omega$

The internal resistance of the second battery $r_2=2\Omega$

Resistance of the resistor $R=10\Omega$

Step 2: Formula Used

Equivalent resistance $\begin{array}{l}{E}_{eq}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}\end{array}$

Ohm's law, $V=IR$

Step 3: Calculate Equivalent EMF

We know that the equivalent resistance is given as,

$$\begin{gathered} \begin{array}{l}{E}_{eq}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}\end{array} \\ =\begin{array}{l}\frac{\frac{12}{1}+\frac{13}{2}}{\frac{1}{1}+\frac{1}{2}}\end{array} \end{gathered}$$

$$\begin{gathered} =\frac{24+13}{2+1} \\ =\frac{37}{3} \end{gathered}$$

Step 4: Calculate Equivalent Resistance

Their equivalent internal resistance is given as,

$$\begin{gathered} \frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2} \\ =1+\frac{1}{2} \\ =\frac{3}{2} \\ \Rightarrow r_{eq}=\frac{2}{3}\Omega \end{gathered}$$

Therefore, $r_{eq}$ and $R$ resistance is present in series, which gives the total equivalent resistance as,

$$\begin{gathered} R_{eq}=R+r_{eq} \\ =10+\frac{2}{3} \\ =\frac{32}{3}\Omega \end{gathered}$$

Step 5: Calculate the Voltage

According to Ohm's law,

$$\begin{gathered} I=\frac{E_{eq}}{R_{eq}} \\ =\frac{{\displaystyle \frac{37}{3}}}{{\displaystyle \frac{32}{3}}} \\ =\frac{37}{32}A \end{gathered}$$

Therefore, the voltage across the resistance $R$,

$$\begin{gathered} V=IR \\ =\frac{37}{32}\times 10=11.5V \end{gathered}$$

Hence, the correct answer is option (D).