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Test: Cells, EMF & Resistance (NCERT)


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Test: Cells, EMF & Resistance (NCERT) - Question 1

A cell having an emf ε and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by

Detailed Solution for Test: Cells, EMF & Resistance (NCERT) - Question 1

image
Current in the circuit,

Potential difference across R,


When R = 0,V = 0
R = ∞,V = ε

Test: Cells, EMF & Resistance (NCERT) - Question 2

A battery of emf 15V and internal resistance of 4Ω is connected to a resistor. If the current in the circuit is 2A and the circuit is closed. Resistance of the resistor and terminal voltage of the battery will be

Detailed Solution for Test: Cells, EMF & Resistance (NCERT) - Question 2

Given, ε = 15V,
r = 4Ω,
I = 2A
Now, for resistance of the resistors
ε − Ir = V = IR;
15 − 2 × 4=2 × R;
15 − 8  = 2R
R = 7/2 = 3.5Ω.
Terminal voltage of battery,
V = IR = 2 × 3.5 = 7V

Test: Cells, EMF & Resistance (NCERT) - Question 3

The battery of a trunk has an emf of 24V. If the internal resistance of the battery is 0.8Ω. What is the maximum current that can be drawn from the battery?

Detailed Solution for Test: Cells, EMF & Resistance (NCERT) - Question 3

Here, ε = 24V and r = 0.8Ω
For the maximum current from the battery.
ε = Ir   (∵ R = 0)
∴ I = ε / r = 24/0.8 = 30A

Test: Cells, EMF & Resistance (NCERT) - Question 4

A battery having 12V emf and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 1A, then the resistance of resistor and lost voltage of the battery when circuit is closed will be

Detailed Solution for Test: Cells, EMF & Resistance (NCERT) - Question 4

Here,
ε = 12V,  r = 3Ω, I = 1A, V = IR = ε − Ir
∴ 
= 12 − 3 = 9Ω
and V = IR = 1 × 9 = 9V

Test: Cells, EMF & Resistance (NCERT) - Question 5

When a current of 2A flows in a battery from negative to positive terminal, the potential difference across it is 12V. If a current of 3A flowing in the opposite direction produces a potential difference of 15V, the emf of the battery is

Detailed Solution for Test: Cells, EMF & Resistance (NCERT) - Question 5

Let ε be emf and r be internal resistance of the battery.
In first case,
12 = ε − 2r ...(i)
In second case,
15 = ε + 3r ...(ii)
Subtract (i) from (ii), we get
r  = 3 / 5Ω
Putting this value of r in eqn. (i), we get


= 66/5 = 13.2V

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