Question 1

A capacitor of capacitance C takes 2 s to reach 63% of its maximum charge when connected in series to a resistance R and a battery of emf. How long does it take (from zero initial charge) for this capacitor to reach 95% of its maximum charge?

Accepted Answer

The answer of A capacitor of capacitance C takes 2...

Question 2

A galvanometer has an internal resistance of 100 Ω and deflects full-scale at 2.00 mA. What size resistor should be added to it to convert it to a millivoltmeter capable of reading up to 400 mV?

Accepted Answer

The answer of A galvanometer has an internal resistance of...

Question 3

A galvanometer has a coil with a resistance of 24.0 Ω. A current of 180 μA causes full-scale deflection. If the galvanometer is to be used to construct an ammeter that deflects full scale for 10.0 A, what shunt resistor is required?

Accepted Answer

The answer of A galvanometer has a coil with a...

Question 4

A 2.0-μF capacitor is charged to 12 V and then discharged through a 4.0 × 10⁶ Ω resistor. How long will it take for the voltage across the capacitor to drop to 3.0 V?

Accepted Answer

The time constant τ for an RC circuit is given by τ = R × C. Here, τ = 4.0 × 10^6 Ω × 2.0 × 10^-6 F = 8 s. The voltage across a discharging capacitor is given by V(t) = V0 * e^(-t/τ). Solving 3 = 12 * e^(-t/8) gives t ≈ 11 s.

Question 5

An analog ammeter able to have a 0.5 A maximum reading is to be built using a small coil of resistance R = 20 Ω, and full scale deflection at 1 mA. What resistance should be added to this coil and how should it be connected to the coil?

Accepted Answer

The answer of An analog ammeter able to have a...

Question 6

A galvanometer with a coil resistance of 40.0 Ω deflects full scale for a current of 2.0 mA. What series resistance should be used with this galvanometer in order to construct a voltmeter that deflects full scale for 50 V?

Accepted Answer

To convert the galvanometer into a voltmeter that reads full scale at 50 V, we need to calculate the series resistance required. The full-scale current is 2.0 mA, and the coil resistance is 40 Ω. Using Ohm's law (V = IR), the total resistance needed is 50 V / 0.002 A = 25,000 Ω. Subtracting the coil resistance (40 Ω) from this gives the series resistance: 25,000 Ω - 40 Ω = 24,960 Ω, which is approximately 25 kΩ.

Question 7

A small galvanometer coil of resistance R = 20 Ω and full-scale deflection at 1 mA is connected in series with a 4980 Ω resistance to build an analog voltmeter. What is the maximum voltage that this voltmeter can read?

Accepted Answer

The answer of A small galvanometer coil of resistance R...

Question 8

A 4.0-μF capacitor is charged to 6.0-V. It is then connected in series with a 3.0-MΩ resistor and connected to a 12-V battery. How long after being connected to the battery will the voltage across the capacitor be 9.0 V?

Accepted Answer

The time constant τ for an RC circuit is given by τ = R*C. Here, R = 3.0 MΩ and C = 4.0 μF, so τ = 3.0 * 10^6 * 4.0 * 10^-6 = 12 s. The voltage across the capacitor as a function of time t is given by V(t) = V_final + (V_initial - V_final) * e^(-t/τ). Solving for t when V(t) = 9.0 V, V_initial = 6.0 V, V_final = 12.0 V, and τ = 12 s, we find t ≈ 8.3 s.

Question 9

A galvanometer has an internal resistance of 100 Ω and deflects full-scale at 2.00 mA. What size resistor should be added to it to convert it to a milliammeter capable of reading up to 4.00 mA?

Accepted Answer

To convert the galvanometer to a milliammeter that reads up to 4.00 mA, a shunt resistor is needed in parallel. The shunt resistor should allow 2.00 mA to pass through it when the total current is 4.00 mA. Using the formula for parallel resistors and the current division rule, the shunt resistor should be 100 Ω.

Question 10

A 4.0-MΩ resistor is connected in series with a 0.50-μF capacitor. The capacitor is initially uncharged. The RC combination is charged by a 9.0-V battery. What is the change in voltage between t = RC and t = 3RC?

Accepted Answer

The voltage across a charging capacitor in an RC circuit is given by V(t) = V0(1 - e^(-t/RC)). At t = RC, V(RC) = 9(1 - e^(-1)) ≈ 5.7 V. At t = 3RC, V(3RC) = 9(1 - e^(-3)) ≈ 8.1 V. The change in voltage is 8.1 V - 5.7 V = 2.4 V, which is closest to 2.9 V.

Quiz 26: Dc Circuits