Q1: When the current in a pocket calculator is 0.0002 A, how much charge flows every minute?
Given:
Current, I = 0.0002 A = 2 × 10⁻⁴ A
Time, t = 1 minute = 60 seconds
Formula: I = Q / t
Q = I × t
Calculation:
Q = (2 × 10⁻⁴) × 60
Q = 120 × 10⁻⁴
Q = 0.012 C
Q = 12 × 10⁻³ C = 12 mC
Answer: 12 mC
Current, I = 0.0002 A = 2 × 10⁻⁴ A
Time, t = 1 minute = 60 seconds
Formula: I = Q / t
Q = I × t
Calculation:
Q = (2 × 10⁻⁴) × 60
Q = 120 × 10⁻⁴
Q = 0.012 C
Q = 12 × 10⁻³ C = 12 mC
Answer: 12 mC
Q2: Calculate the amount of current that an electric heater uses to heat a room in 5 minutes if the charge is 2100 C.
Given:
Charge, Q = 2100 C
Time, t = 5 minutes = 5 × 60 = 300 seconds
Formula: I = Q / t
Calculation:
I = 2100 / 300
I = 7 A
Answer: 7 A
Charge, Q = 2100 C
Time, t = 5 minutes = 5 × 60 = 300 seconds
Formula: I = Q / t
Calculation:
I = 2100 / 300
I = 7 A
Answer: 7 A
Q3: A potential difference of 90 V exists between two points. The amount of work done when an unknown charge is moved between the points is 450 J. Determine the charge amount.
Given:
Potential difference, V = 90 V
Work done, W = 450 J
Formula: V = W / Q
Q = W / V
Calculation:
Q = 450 / 90
Q = 5 C
Answer: 5 C
Potential difference, V = 90 V
Work done, W = 450 J
Formula: V = W / Q
Q = W / V
Calculation:
Q = 450 / 90
Q = 5 C
Answer: 5 C
Q4: Calculate the potential difference between two points A and B if it takes 9 × 10⁻⁴ J of external work to move a charge of +9 μC from A to B.
Given:
Work done, W = 9 × 10⁻⁴ J
Charge, Q = 9 μC = 9 × 10⁻⁶ C
Formula: V = W / Q
Calculation:
V = (9 × 10⁻⁴) / (9 × 10⁻⁶)
V = 10²
V = 100 V
Answer: 100 V
Work done, W = 9 × 10⁻⁴ J
Charge, Q = 9 μC = 9 × 10⁻⁶ C
Formula: V = W / Q
Calculation:
V = (9 × 10⁻⁴) / (9 × 10⁻⁶)
V = 10²
V = 100 V
Answer: 100 V
Q5: The potential difference applied to a portable radio terminal is 6.0 Volts. Determine the resistance of the radio when a current of 20 mA flows through it.
Given:
Potential difference, V = 6.0 V
Current, I = 20 mA = 20 × 10⁻³ A = 0.02 A
Formula: Ohm's Law: V = IR
R = V / I
Calculation:
R = 6.0 / 0.02
R = 300 Ω
Answer: 300 Ω
Potential difference, V = 6.0 V
Current, I = 20 mA = 20 × 10⁻³ A = 0.02 A
Formula: Ohm's Law: V = IR
R = V / I
Calculation:
R = 6.0 / 0.02
R = 300 Ω
Answer: 300 Ω
Q6: Resistances of 4 Ω, 6 Ω, and 12 Ω are connected in parallel and then connected to a 6V emf source. Determine the value of
i. The circuit's equivalent resistance.
ii. The total current flowing through the circuit.
iii. The current that flows through each resistance.
i. The circuit's equivalent resistance.
ii. The total current flowing through the circuit.
iii. The current that flows through each resistance.
Given:
R₁ = 4 Ω
R₂ = 6 Ω
R₃ = 12 Ω
Voltage, V = 6 V
All connected in parallel
Formula for parallel combination:
1/Rₑq = 1/R₁ + 1/R₂ + 1/R₃
Calculation (i) Equivalent resistance:
1/Rₑq = 1/4 + 1/6 + 1/12
1/Rₑq = 0.25 + 0.1667 + 0.0833
1/Rₑq = 0.5
Rₑq = 1/0.5 = 2 Ω
Calculation (ii) Total current:
Iₜ = V / Rₑq = 6 / 2 = 3 A
Calculation (iii) Current through each resistance:
I₁ = V / R₁ = 6 / 4 = 1.5 A
I₂ = V / R₂ = 6 / 6 = 1 A
I₃ = V / R₃ = 6 / 12 = 0.5 A
Answer:
i. Equivalent resistance = 2 Ω
ii. Total current = 3 A
iii. Currents: I₁ = 1.5 A, I₂ = 1 A, I₃ = 0.5 A
R₁ = 4 Ω
R₂ = 6 Ω
R₃ = 12 Ω
Voltage, V = 6 V
All connected in parallel
Formula for parallel combination:
1/Rₑq = 1/R₁ + 1/R₂ + 1/R₃
Calculation (i) Equivalent resistance:
1/Rₑq = 1/4 + 1/6 + 1/12
1/Rₑq = 0.25 + 0.1667 + 0.0833
1/Rₑq = 0.5
Rₑq = 1/0.5 = 2 Ω
Calculation (ii) Total current:
Iₜ = V / Rₑq = 6 / 2 = 3 A
Calculation (iii) Current through each resistance:
I₁ = V / R₁ = 6 / 4 = 1.5 A
I₂ = V / R₂ = 6 / 6 = 1 A
I₃ = V / R₃ = 6 / 12 = 0.5 A
Answer:
i. Equivalent resistance = 2 Ω
ii. Total current = 3 A
iii. Currents: I₁ = 1.5 A, I₂ = 1 A, I₃ = 0.5 A
Q7: A 220 V circuit is used to power two 120 watt and 80 watt light bulbs. Which bulb has the greater resistance and which one has the higher current?
Given:
Voltage, V = 220 V
Power of bulb 1, P₁ = 120 W
Power of bulb 2, P₂ = 80 W
Formulas:
P = V² / R → R = V² / P
P = VI → I = P / V
Calculation for resistance:
R₁ = (220)² / 120 = 48400 / 120 = 403.33 Ω
R₂ = (220)² / 80 = 48400 / 80 = 605 Ω
80 W bulb has greater resistance (605 Ω > 403.33 Ω)
Calculation for current:
I₁ = 120 / 220 = 0.545 A
I₂ = 80 / 220 = 0.364 A
120 W bulb has higher current (0.545 A > 0.364 A)
Answer:
Greater resistance: 80 W bulb
Higher current: 120 W bulb
Voltage, V = 220 V
Power of bulb 1, P₁ = 120 W
Power of bulb 2, P₂ = 80 W
Formulas:
P = V² / R → R = V² / P
P = VI → I = P / V
Calculation for resistance:
R₁ = (220)² / 120 = 48400 / 120 = 403.33 Ω
R₂ = (220)² / 80 = 48400 / 80 = 605 Ω
80 W bulb has greater resistance (605 Ω > 403.33 Ω)
Calculation for current:
I₁ = 120 / 220 = 0.545 A
I₂ = 80 / 220 = 0.364 A
120 W bulb has higher current (0.545 A > 0.364 A)
Answer:
Greater resistance: 80 W bulb
Higher current: 120 W bulb
