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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
Current in the circuit,
Potential difference across R,
When R = 0,V = 0
R = ∞,V = ε
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
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
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?
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
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
Here,
ε = 12V, r = 3Ω, I = 1A, V = IR = ε − Ir
∴
= 12 − 3 = 9Ω
and V = IR = 1 × 9 = 9V
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
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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