SATHEE: current-electricity Question 20

Current Electricity

current-electricity Question 20

Question: Q. 8. Two cells of emf $E_{1}$ and $E_{2}$ have internal resistance $r_{1}$ and $r_{2}$ . Deduce an expression for equivalent emf of their parallel combination. $◻$ [SQP II 2017]

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Solution:

Ans.

$\begin{aligned} I = & I_{1} + I_{2} = & \frac{E_{1} - V}{r_{1}} + \frac{E_{2} - V}{r_{2}} I = & (\frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}}) - V (\frac{1}{r_{1}} + \frac{1}{r_{2}}) & I = \frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} r_{2}} - V (\frac{r_{1} + r_{2}}{r_{1}}) \end{aligned}$

or $V = (\frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} r_{2}}) \times \frac{r_{1} r_{2}}{r_{1} + r_{2}}) - I {(\frac{r_{1} r_{2}}{r_{1} + r_{2}})}_{1 / 2}$

or $V = (\frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} + r_{2}}) - I (\frac{r_{1} r_{2}}{r_{1} + r_{2}})$

Comparing above with

$V = E_{e q} - I_{e q} \times R_{e q}$

We get $E_{e q} = \frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} + r_{2}}$

AI Q. 9. A cell of emf $4 V$ and internal resistance $1 Ω$ is connected to a d.c. source of $10 V$ through a resistor of $5 Ω$ . Calculate the terminal voltage across the cell during charging.

U [O.D. III 2017]

Ans. Calculation of Current

Calculation of Terminal Voltage

$10 - 4 = I (1 + 5)$

$∴ I = 1 A$ $∴$ Terminal voltage across cell

$\begin{aligned} = (4 + 1 \times 1) V & = 5 V \end{aligned}$

[CBSE Marking Scheme 2017]

Current Electricity

current-electricity Question 20

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Current Electricity

current-electricity Question 20

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

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NCERT Exercise Video Solution

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