Question 488: Current Electricity

Question

Two batteries of e.m.f 4 V and 8 V with internal resistance $1 \Omega$ and $2 \Omega$ respectively are connected in a circuit with a resistance of $9 \Omega$ as shown in the figure. The current and potential difference between the points ' P ' and ' Q ' is $\mathrm{R}=9 \Omega$

Accepted Answer

Applying Kirchhoff's voltage law to the Given loop QPQ ,
$$ \begin{aligned} &-2 \mathrm{I}+8-4-(1 	imes \mathrm{I})-9 \mathrm{I}=0 \Rightarrow \mathrm{I}=\frac{1}{3} \mathrm{~A}\ &	herefore \quad 	ext { Potential difference across } \mathrm{PQ}=\frac{1}{3} 	imes 9=3 \mathrm{~V} \end{aligned} $$

Answer

$\frac{1}{3} \mathrm{~A}$ and 4 V

Answer

$\frac{1}{3}$ A and 3 V

Answer

$\frac{1}{2}$ A and 5 V

Answer

$\frac{1}{6} \mathrm{~A}$ and 3 V

Current Electricity

Two batteries of e.m.f 4 V and 8 V with internal resistance 1 \Omega and 2 \Omega respectively are connected in a circuit with a resistance of 9 \Omega as shown in the figure. The current and potential difference between the points ' P ' and ' Q ' is \mathrm{R}=9 \Omega

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