SATHEE: current-electricity Question 15

Current Electricity

current-electricity Question 15

Question: Q. 19. A $10 V$ battery of negligible internal resistance is connected across a $200 V$ battery and a resistance of $38 Ω$ as shown in the figure. Find the value of the current in the circuit.

U] [Delhi I, II, III 2013, 2018]

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Solution:

Ans. From the circuit we see that, the two batteries are joined together with positive terminal, so equivalent emf of both the batteries will result as :

$\begin{matrix} (1) & I = \frac{E}{R} = \frac{190}{38} = 5 A \end{matrix}$

[CBSE Marking Scheme 2013, 2018]

Alternatively :

$\begin{aligned} E_{net} & = E_{1} - E_{2} & = (200 - 10) = 190 V \end{aligned}$

Current in circuit, $I = \frac{E_{net}}{R}$

$= \frac{190}{38} = 5 A$

AI Q. 20. Two identical cells, each of emf $E$ , having negligible internal resistance, are connected in parallel with each other across an external resistance $R$ . What is the current through this resistance?

R [O.D. I, II, III 2013]

Since internal resistance is negligible, so total resistance $= R$

Potential difference across the resistance $= E$ .

Hence, current through the resistance in the circuit :

$I = \frac{E}{R}$

[CBSE Marking Scheme 2013]

Current Electricity

current-electricity Question 15

Short Answer Type Questions-I

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Current Electricity

current-electricity Question 15

Short Answer Type Questions-I

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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