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
(b) Current flowing through the circuit = I
Emf of the battery, E = 12 V
Total resistance of the circuit, R = 6 Ω
Applying Ohm’s law is, I = E/R
= 12/6 = 2 A
Potential drop across 1 Ω resistor = V1
Using Ohm’s law, the value of V1 can be obtained as
V1 = 2 × 1= 2 V … (I)
Potential drop across 2 Ω resistor = V2
Using Ohm’s law, the value of V2 can be determined as
V2 = 2 × 2= 4 V … (II)
Potential drop across 3 Ω resistor = V3
Using Ohm’s law, the value of V3 is
V3 = 2 × 3= 6 V … (III)
∴ the potential drop across 1 Ω, 2 Ω, and 3 Ω resistors are 2 V, 4 V, and 6 V respectively.
Q4: (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
Answer: (a) Given three resistors connected in parallel ie. R1 = 2 Ω, R2 = 4 Ω, and R3 = 5 Ω
For parallel combiation, total effective resistance (R) of the combination is:
1/ R = 1/ R1 + 1/ R2 + 1/ R3 = 1/2 + 1/4 + 1/5 = 19/20
⇒ R = 20/19 Ω ........Answer
(b) Emf of the battery, V = 20 V
Current (I1) flowing through resistor R1 is = V1/R = 20/2 = 10 A
Current (I2) flowing through resistor R2 is = V2/R = 20/4 = 5 A
Current (I3) flowing through resistor R3 is = V3/R = 20/5 = 4 A
Total current, I = I1 + I2 + I3 = 10 + 5 + 4 = 19 A
Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the total current is 19 A.
Q5: At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10 -4°C-1
Answer: Room temperature, T = 27°C
Resistance of the heating element at T, R = 100 Ω
Let T1 is the increased temperature of the filament.
Resistance of the heating element at T1, R1 = 117 Ω
Temperature co-efficient of the material of the filament, α = 1.70 × 10 -4°C-1
Since α = (R1 - R)/ (T-T1)R
⇒ T-T1 = (R1 - R)/αR
⇒ T1 - 27 = 1000
⇒ T1 = 1027°C
Thus at 1027°C, the resistance of the element is 117Ω.
(In progress...)
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