Question 1

A simple circuit has a total resistance of 30 Ω. If a 2.0-A current is maintained in this circuit, how much energy is dissipated in this circuit in 4.0 seconds?

Accepted Answer

Using Ohm's Law, we can find the voltage in the circuit:

V = IR = (2.0 A)(30 Ω) = 60 V

The power dissipated in the circuit is:

P = IV = (2.0 A)(60 V) = 120 W

The energy dissipated in 4.0 seconds is:

E = Pt = (120 W)(4.0 s) = 480 J

Therefore, the answer is choice D) 480 J.

Question 2

FIGURE 21-3 
-A 4-Ω resistor is connected in parallel with a 12-Ω resistor and this combination is connected to a DC power supply with voltage V as shown in Figure 21-3. If the total current in this circuit is 2 A, what is the value of voltage V?

Accepted Answer

The equivalent resistance of the parallel combination of 4 Ω and 12 Ω resistors is 3 Ω. Using Ohm's Law, we have V = IR = (2 A)(3 Ω) = 6 V. Therefore, the answer is (C) 6 V.

Question 3

A 200-W light bulb is connected across 110 V. What current will flow through this bulb?

Accepted Answer

The answer of A 200-W light bulb is connected across...

Question 4

FIGURE 21-8 
-Five equal resistors, of value 2.0 Ω each, are connected as shown in Figure 21-8. What is the equivalent resistance of this circuit?

Accepted Answer

The answer of FIGURE 21-8 
-Five equal resistors, of value...

Question 5

FIGURE 21-4 
-A 100 V DC signal is applied to four resistors as shown in Figure 21-4. The values of the resistors are 20 Ω, 40 Ω, 60 Ω, and 80 Ω. What is the voltage across the 40 Ω resistor?

Accepted Answer

The resistors are in series, so the same current flows through each. The total resistance is 200 Ω (20+40+60+80). The current is 100 V / 200 Ω = 0.5 A. The voltage across the 40 Ω resistor is 0.5 A * 40 Ω = 20 V.

Question 6

A 100-W light bulb is connected to a 110-V source. What current flows through the lamp?

Accepted Answer

We can use Ohm's Law to find the current:
I = V/R
where V is the voltage and R is the resistance.

Power is also related to voltage and current by:
P = IV

We can rearrange this equation to solve for resistance:
R = P / I^2

Substituting in the values given:
P = 100 W
V = 110 V

R = (100 W) / I^2

The resistance of the light bulb is constant, so we can use this to solve for the current:

110 V = I R
I = 110 V / R
I = 110 V / (100 W / I^2)
I = 1.1 A

Therefore, the answer is E) 0.91 A.

Question 7

A 100-W light bulb is connected to a 110-V source. What is the resistance of this bulb?

Accepted Answer

The resistance can be calculated using the formula R = V²/P, where V is voltage and P is power. Substituting the given values, R = (110)² / 100 = 121 Ω.

Question 8

FIGURE 21-6 
-Three resistors of values 2 Ω, 6 Ω and 12 Ω are connected across a 2.0-V DC voltage source as shown in Figure 21-6. What is the total current I flowing this circuit?

Accepted Answer

Using Ohm's Law, we can determine the individual currents in each resistor:
- The current through the 2 Ω resistor is I = V/R = 2/2 = 1 A
- The current through the 6 Ω resistor is I = V/R = 2/6 = 1/3 A
- The current through the 12 Ω resistor is I = V/R = 2/12 = 1/6 A
The total current I flowing through the circuit is the sum of these individual currents:
I = 1 + 1/3 + 1/6 = 1.5 A

Question 9

The resistivity of a 1.0 m long wire is 1.72 × 10^-8 Ωm and its cross sectional area is 2.0 × 10^-6 m². If the wire carries a current of 0.20 A, what is the voltage across the wire?

Accepted Answer

The resistance of the wire can be calculated using the formula:

R = ρL/A

where ρ is the resistivity, L is the length of wire, and A is the cross-sectional area.

R = (1.72 × 10^-8 Ωm)(1.0 m)/(2.0 × 10^-6 m^2) = 8.6 Ω

Using Ohm's Law (V = IR), we can find the voltage across the wire:

V = IR = (0.20 A)(8.6 Ω) = 1.72 V

However, the question asks for the voltage in millivolts (mV), so we need to convert:

1.72 V = 1720 mV

Therefore, the answer is 1.7 mV, which is choice C.

Question 10

FIGURE 21-9 
-A 4.0-Ω resistor is connected in parallel with a 12-Ω resistor and both of these are connected to a DC power supply with voltage V as shown in Figure 21-9. If the total current in this circuit is 2.0 A, what is the current through the 4.0-Ω resistor?

Accepted Answer

The total current is divided between the resistors in parallel. The current through the 4.0-Ω resistor is calculated using the formula I = V/R. The voltage across both resistors is the same, and the current through the 4.0-Ω resistor is higher due to its lower resistance. Using the formula for parallel resistors and the given total current, the current through the 4.0-Ω resistor is 1.5 A.

Question 11

The power rating of a resistor is 0.80 W. If the value of the resistor is 400 Ω, what is the maximum current it can draw?

Accepted Answer

The maximum current that a resistor can draw is given by the square root of the power rating divided by the resistance. In this case, the maximum current is:I = sqrt(P/R) = sqrt(0.80 W / 400 Ω) = 0.045 A = 45 mA

Question 12

Four resistors of 20 Ω, 40 Ω, 60 Ω, and 80 Ω are connected across a DC voltage source. If the current through this circuit is 0.5 A, what is the voltage applied to this circuit?

Accepted Answer

The resistors are in series, so the total resistance is 20 + 40 + 60 + 80 = 200 Ω. Using Ohm's Law (V = IR), the voltage is 0.5 A * 200 Ω = 100 V.

Question 13

Four resistors of 20 Ω, 40 Ω, 60 Ω, and 80 Ω are connected across a 50-V DC source. What is the current through this circuit?

Accepted Answer

The total resistance of the circuit is 20 + 40 + 60 + 80 = 200 Ω. The current through the circuit is therefore I = V/R = 50 V / 200 Ω = 0.25 A.

Question 14

FIGURE 21-7 
-Three 2.0-Ω resistors are connected across the sides of an equilateral triangle ABC as shown in Figure 21-7. What is the equivalent resistance between any two points, AB, BC, or AC, of this circuit?

Accepted Answer

The resistors form a delta (Δ) configuration. The equivalent resistance between any two points in a delta configuration of equal resistors is given by R_eq = R/3, where R is the resistance of each resistor. Here, R = 2.0 Ω, so R_eq = 2.0 Ω / 3 = 0.67 Ω. However, the correct calculation involves using the formula for two resistors in parallel (R_parallel = R/2) and adding the third resistor in series, resulting in R_eq = 1.3 Ω.

Question 15

The power rating of a 400 Ω resistor is 0.25 W. What is the maximum voltage?

Accepted Answer

The maximum voltage can be calculated using the formula P = V^2 / R. Rearranging gives V = sqrt(P * R). Substituting the given values, V = sqrt(0.25 W * 400 Ω) = 10 V.

Question 16

A light bulb operating at a voltage of 120 V has a resistance of 200 Ω. What is the power?

Accepted Answer

Using the formula: P = V^2/R, we can find the power (P) in watts.

P = (120 V)^2 / 200 Ω
P = (14400 V^2) / 200 Ω
P = 72 W

Therefore, the power of the light bulb is 72 W. The answer is C.

Question 17

The power rating of a resistor is 0.800 W. If the value of the resistor is 400 Ω, what is the maximum voltage?

Accepted Answer

The maximum voltage can be calculated using the formula P = V^2 / R. Rearranging gives V = sqrt(P * R). Substituting the given values, V = sqrt(0.800 W * 400 Ω) = 17.9 V.

Question 18

A 100-V DC signal is applied to a series circuit composed of four equal resistors 10 Ω each. What is the voltage across each resistor?

Accepted Answer

The answer of A 100-V DC signal is applied to...

Question 19

A light bulb operating at a voltage of 120 Volt has a power of 60 W. What is the current flowing through this bulb?

Accepted Answer

Using the formula P = IV, we can rearrange to solve for current: I = P/V. 
Plugging in the values given: I = 60 W / 120 V = 0.5 A. Therefore, the correct answer is C.

Question 20

FIGURE 21-5 
-Three resistors of values 2 Ω, 6 Ω and 12 Ω are connected across a DC voltage source as shown in Figure 21-5. If the total current through the circuit is 2.0 A, what is the applied voltage?

Accepted Answer

To find the applied voltage, we can use Ohm's Law: V = IR. 
The total resistance of the circuit, RT, is the sum of the individual resistances: RT = 2 Ω + 6 Ω + 12 Ω = 20 Ω. 
Then, we can use Ohm's Law to find the voltage: V = IR = (2.0 A)(20 Ω) = 40 V. 
Therefore, the applied voltage is 40 V.

Quiz 21: Electric Current and Direct-Current Circuits

A simple circuit has a total resistance of 30 Ω. If a 2.0-A current is maintained in this circuit, how much energy is dissipated in this circuit in 4.0 seconds?

FIGURE 21-3 -A 4-Ω resistor is connected in parallel with a 12-Ω resistor and this combination is connected to a DC power supply with voltage V as shown in Figure 21-3. If the total current in this circuit is 2 A, what is the value of voltage V?

A 200-W light bulb is connected across 110 V. What current will flow through this bulb?

FIGURE 21-8 -Five equal resistors, of value 2.0 Ω each, are connected as shown in Figure 21-8. What is the equivalent resistance of this circuit?

FIGURE 21-4 -A 100 V DC signal is applied to four resistors as shown in Figure 21-4. The values of the resistors are 20 Ω, 40 Ω, 60 Ω, and 80 Ω. What is the voltage across the 40 Ω resistor?

A 100-W light bulb is connected to a 110-V source. What current flows through the lamp?

A 100-W light bulb is connected to a 110-V source. What is the resistance of this bulb?

FIGURE 21-6 -Three resistors of values 2 Ω, 6 Ω and 12 Ω are connected across a 2.0-V DC voltage source as shown in Figure 21-6. What is the total current I flowing this circuit?

The resistivity of a 1.0 m long wire is 1.72 × 10-8 Ωm and its cross sectional area is 2.0 × 10-6 m2. If the wire carries a current of 0.20 A, what is the voltage across the wire?

FIGURE 21-9 -A 4.0-Ω resistor is connected in parallel with a 12-Ω resistor and both of these are connected to a DC power supply with voltage V as shown in Figure 21-9. If the total current in this circuit is 2.0 A, what is the current through the 4.0-Ω resistor?