StudentAidUnit 13: Current Electricity — Numericals
12th Class Physics · Unit 13: Current Electricity
13.1.1.0 × 10^7 electrons pass through a conductor in 1.0 μs. Find the current in ampere flowing through the conductor. Electronic charge is 1.6 × 10^-19 C.
Given
Number of electrons
N = 1.0 × 107
Time
Δt = 1.0 μs = 1.0 × 10-6 s
Charge on electron
q = 1.6 × 10-19 C
Formula
Definition of current
I = ΔQΔt
Total charge
ΔQ = N × q
Working
Calculate total charge
ΔQ = 1.0 × 107 × 1.6 × 10-19
= 1.6 × 10-12 C
Calculate current
I = frac{1.6 × 10-12{1.0 × 10-6
= 1.6 × 10-6 A
I = 1.6 × 10-6 A
13.2.0.75 A current flows through an iron wire when a battery of 1.5 V is connected across its ends. The length of the wire is 5.0 m and its cross sectional area is 2.5 × 10^-7 m^2. Compute the resistivity of iron.
Given
Current
I = 0.75 A
Potential difference
V = 1.5 V
Length of wire
L = 5.0 m
Area of wire
A = 2.5 × 10-7 m2
Formula
Resistance from Ohm's law
R = VI
Working
Calculate resistance
R = 1.50.75
= 2.0 Ω
Formula
Resistivity formula
rho = R × AL
Working
Calculate resistivity
rho = frac{2.0 × 2.5 × 10-7{5.0}
= 1.0 × 10-7 Ω·m
rho = 1.0 × 10-7 Ω·m
13.3.A platinum wire has resistance of 10 Ω at 0°C and 20 Ω at 273°C. Find the value of temperature coefficient of resistance of platinum.
Given
Resistance at 0°C
R0 = 10 Ω
Resistance at 273°C
Rt = 20 Ω
Temperature at 0°C
t0 = 0°C + 273 = 273 K
Temperature at 273°C
t = 273°C + 273 = 546 K
Temperature difference
Δt = t - t0 = 546 - 273 = 273 K
Formula
Temperature coefficient
α = frac{Rt - R0{R0 × Δt}
Working
Substituting values
α = 20 - 1010 × 273
= 102730
= 1273 K-1
= 3.66 × 10-3 K-1
α = 3.66 × 10-3 K-1
13.4.The potential difference between the terminals of a battery in open circuit is 2.2 V. When it is connected across a resistance of 5.0 Ω, the potential falls to 1.8 V. Calculate the current and the internal resistance of the battery.
Given
EMF of battery
E = 2.2 V
External resistance
R = 5.0 Ω
Terminal voltage under load
V = 1.8 V
Formula
Ohm's law for external circuit
I = VR
Working
Calculate current
I = 1.85.0
= 0.36 A
Formula
Battery equation
E = V + Ir
Rearranging
E - V = Ir
Internal resistance
r = E - VI
Working
Calculate internal resistance
r = 2.2 - 1.80.36
= 0.40.36
= 1.11 Ω
I = 0.36 A, quad r = 1.11 Ω
13.5.Calculate the currents in the three resistances of the circuit shown in figure.
Given
Resistance 1
R1 = 10 Ω
Resistance 2
R2 = 30 Ω
Resistance 3
R3 = 15 Ω
EMF 1
E1 = 40 V
EMF 2
E2 = 60 V
EMF 3
E3 = 50 V
Formula
Kirchhoff's 2nd law, loop 1
-E1 - I1R1 - (I1 - I2)R2 + E2 = 0
Expanding
-40 - 10I1 - 30(I1 - I2) + 60 = 0
-40 - 10I1 - 30I1 + 30I2 + 60 = 0
20 - 40I1 + 30I2 = 0
Divide by 10
2 - 4I1 + 3I2 = 0 quad cdots (i)
Kirchhoff's 2nd law, loop 2
-E2 - (I2 - I1)R2 - I2R3 + E3 = 0
Expanding
-60 - (I2 - I1) × 30 - I2 × 15 + 50 = 0
-60 - 30I2 + 30I1 - 15I2 + 50 = 0
-10 + 30I1 - 45I2 = 0
Divide by 5
-2 - 9I2 + 6I1 = 0 quad cdots (ii)
Multiply equation (i) by 3
6 - 12I1 + 9I2 = 0
Add to equation (ii)
6 - 12I1 + 9I2 - 2 - 9I2 + 6I1 = 0
4 - 6I1 = 0
6I1 = 4
I1 = 23 = 0.66 A
Put I₁ in equation (ii)
-2 - 9I2 + 6 × 23 = 0
-2 - 9I2 + 4 = 0
2 - 9I2 = 0
I2 = 29 = 0.22 A
Current through R₂
I3 = I1 - I2 = 0.66 - 0.22 = 0.44 A
I1 = 0.66 A, quad I2 = 0.44 A, quad I3 = 0.22 A