Question: Q. 2. A $9 \mathrm{~V}$ battery is connected in series with a resistor. The terminal voltage is found to be $8 \mathrm{~V}$. Current through the circuit is measured as $5 \mathrm{~A}$. What is the internal resistance of the battery?
A [CBSE SQP 2018-19]
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Solution:
Ans.
$$ \begin{aligned} r & =\frac{E-V}{I} \ & =\frac{9 \mathrm{~V}-8 \mathrm{~V}}{5 \mathrm{~A}} \ & =0.2 \Omega \end{aligned} $$
[CBSE Marking Scheme 2018]
[AI Q. 3. Two electric bulbs $P$ and $Q$ have their resistances in the ratio of $1: 2$. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs. A [Delhi O.D. 2018]
Sol. Formula
Stating that currents are equal
Ratio of powers
Power $=I^{2} R$
The current, in the two bulbs, is the same as they are connected in series.
$$ \begin{aligned} \therefore \quad \frac{P_{1}}{P_{2}} & =\frac{I^{2} R_{1}}{I^{2} R_{2}}=\frac{R_{1}}{R_{2}} \ & =\frac{1}{2} \end{aligned} $$
[CBSE Marking Scheme 2018]
Detailed Answer :
Since in series combination of resistance, the current flowing is same but voltage is different, therefore power dissipation is given by
$$ P=I^{2} R \Rightarrow P \propto R \Rightarrow \frac{P_{1}}{P_{2}}=\frac{R_{1}}{R} \quad 1 / 2+1 / 2 $$
Now for two bulbs $P$ and $Q$, we have
$$ \frac{P_{1}}{P_{2}}=\frac{R_{1}}{R_{2}}=\frac{1}{2} $$
(Since $R_{1}: R_{2}=1: 2$ given).
$\Rightarrow$ Ratio of Power, $P_{1}: P_{2}=1: 2$
