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Current Electricity

*NCERT Solution *
Q1: The storage battery of a car has
an emf of 12 V. If the internal resistance of the battery
is 0.4Ω, what is the maximum current that can be drawn from the battery?
Answer: Emf of the battery, E = 12 V

Internal resistance of the battery, r = 0.4 Ω

Maximum current drawn from the battery = I

According to Ohm’s law,

E = Ir

I = E/r =
12/0.4 = 30 A

Thus, the maximum current drawn from the given battery is
30 A.

Q2: A battery of emf 10 V and internal
resistance 3 Ω is connected to a resistor. If the current
in the circuit is 0.5 A, what is the resistance of the resistor? What
is the terminal voltage
of the battery when the circuit is closed?
Answer: Given,

Emf of the battery, E = 10 V

Internal resistance of the battery, r = 3 Ω

Current in the circuit, I = 0.5 A

Resistance of the resistor = R

Applying Ohm’s law,

I = E/(R + r)

R + r = E/I = 10/0.5 =
20 Ω

∴ 20 - 3 =
17 Ω

Terminal voltage of the resistor =
V

According to Ohm’s law,

V = IR
= 0.5 × 17
= 8.5 V

∴ the resistance of the resistor
is 17 Ω and the terminal voltage is
8.5 V.

Q3: a) Three resistors 1 Ω, 2 Ω, and 3
Ω are combined in series. What is the total
resistance of the combination?
b) If the combination is connected to a battery of emf 12 V and
negligible internal
resistance, obtain the potential drop across each resistor.
Answer:

(a) Three resistors of resistances 1 Ω, 2 Ω, and 3 Ω
are combined in series.

Total
resistance of the series combination is the algebraic sum of individual
resistances.

⇒ Total resistance = 1
+ 2 + 3 = 6 Ω