SATHEE: current-electricity Question 16

Current Electricity

current-electricity Question 16

Question: Q. 2. A $9 V$ battery is connected in series with a resistor. The terminal voltage is found to be $8 V$ . Current through the circuit is measured as $5 A$ . What is the internal resistance of the battery?

A [CBSE SQP 2018-19]

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Solution:

Ans.

$\begin{aligned} r & = \frac{E - V}{I} & = \frac{9 V - 8 V}{5 A} & = 0.2 Ω \end{aligned}$

[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.

$\begin{aligned} ∴ \frac{P_{1}}{P_{2}} & = \frac{I^{2} R_{1}}{I^{2} R_{2}} = \frac{R_{1}}{R_{2}} & = \frac{1}{2} \end{aligned}$

[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

$P = I^{2} R \Rightarrow P \propto R \Rightarrow \frac{P_{1}}{P_{2}} = \frac{R_{1}}{R} 1 / 2 + 1 / 2$

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, $P_{1} : P_{2} = 1 : 2$

Current Electricity

current-electricity Question 16

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Current Electricity

current-electricity Question 16

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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