3.6: DC Circuits

Question 1

Answer 1

(b) 16V

Question 2

130. Ohms Law may be written as:
     (a) I = V * R
     (b) R = V * I
     (c) V = I * R

Answer 2

(c) V = I * R

Question 3

Answer 3

(a) Not continuous

Question 4

132. The current in a 60 watt, 120 Volt bulb would be:
     (a) 2 Amps
     (b) 0.5 Amps
     (c) 9 Amps

Answer 4

(b) 0.5 Amps

Note: Power = Current * Voltage
60 = 0.5 * 120

Question 5

Answer 5

(c) 30 mA

I = V/R, I = 3/100
I = 3x10^-2 Amps
I = 0.03 Amps or 30mA

Question 6

Answer 6

(a) 8 Amps

Resistors in parallel = (R1R2) / (R1 + R2)
=(515) / (5 + 15) , = 75/20, = 7.5/2
I = V / R , = I = 30V / 7.5/2
Invert and multiply 30 * 2 / 7.5 = 8Amps

Question 7

Answer 7

(a) the algebraic sum of all voltages in a loop of components will be equal to zero

Question 8

Answer 8

(c) 5 Amps

Note: Kirchoff’s current law states that the algebraic sum of all voltages in a loop of components will be equal to zero
Entering Node: 10+1+7 = 18A
Leaving Node: 6+4+3 = 13A, therefore X must be 5A

Question 9

Answer 9

(c) 50 ohms

If the the voltage drop across the bulb is 3V, the voltage drop across the resistor must be 25V to equal supply voltage of 28V
Ohm’s Law, R = V / I. R = 25V / 0.5A, R = 50ohms

Question 10

138. What is the effective source voltage?
     (a) 15 V
     (b) 25 V
     (c) 50 V
     (d) 75 V

Answer 10

(b) 25V

Series Aiding and Opposing Sources:
‘‘In many practicaly applications, a circuit may contain more than one source of EMF. Sources of EMF that cause current to flow in the same direction are considered to be series aiding and the voltages are added.
Sources of EMF that would tend to force current in opposite directions are said to be series opposing, and the effective source voltage is the difference between the opposing voltages. When two opposing sources are inserted into a circuit current flow would be in a direction determined by the larger source.’’

Question 11

Answer 11

(a) Open

Question 12

Answer 12

(a) 10 volts

Total current in circuit:I = V / R, I = 30V / 30 ohms, I = 1A
Voltage drop across resistor, V = I * R, V = 1A * 10ohms, V = 10 volts

Question 13

Answer 13

I = V / R
I = 24V / 2000 ohms
I = (2)(12) V / (2)(1000) ohms …cancel 2’s
I = 12/1000, I = 12x10^-3, 12 mA

Question 14

Answer 14

(a) 1.2 Mega ohms

R = V / I, R = 600 / 0.5x10^-3
R = 1200 / 1/1000
Invert and mulitply, 1200 * 1000 = 1200000 or 1.2 Mega ohms

Question 15

Answer 15

(a) 1 ohms

Rule of Parallel Resistors
1/Rtotal = 1/R1 + 1/R2 + …
R1R2R3R4R5R6R6 / (R1R2R3R4R5R6)7
7^7 / (7^6)*7 , = 7^7 / 7^7 = 1

Question 16

Answer 16

(b) 12 mA

I = V/R, I = 12V/1000, I = 12x10^-3 or 12mA

Question 17

Answer 17

(c) the total current is equal to the sum of the currents through the individual branches of the circuit

Question 18

10. A 5 ohm and a 15 ohm resistor are connected in parallel across a 30V battery. Calculate the
    total current flowing in the circuit.
    (a) 8 Amps
    (b) 6 Amps
    (c) 1.5 Amps

Answer 18

(a) 8 Amps

Question 19

11. If a 3V bulb operating at 0.5 amps is required to be fitted to a 28V circuit. What is the value of
    the resistor required to be fitted in series with the bulb to stop it from blowing.
    (a) 20 ohms
    (b) 35 ohms
    (c) 50 ohms

Answer 19

(c) 50 ohms

Question 20

Answer 20

(b) 25 V

Question 21

13. Which statement is correct in relation to a parallel circuit
    (a) the current is equal in all portions of the circuit
    (b) the current in Amperes is the product of the EMF in Volts times the total resistance of the
    circuit in Ohms
    (c) the total current is equal to the sum of the currents through the individual branches of the
    circuit

Answer 21

(c) the total current is equal to the sum of the currents through the individual branches of the
circuit

Question 22

11. A resistor of 1 KΩ is connected across a 30V supply. The current flowing in the circuit is:
    (a) 0.3Amp
    (b) 3Amps
    (c) 30mA

Answer 22

(c) 30mA

Question 23

12. Kirchhoff’s law states
    a. the algebraic sum of all the current at a junction will be equal to zero
    b. the algebraic sum of all the currents entering or leaving a series of components
    will be equal to one
    c. the inverse sum of all the voltages entering or leaving a series of components
    will be equal to one

Answer 23

a. the algebraic sum of all the current at a junction will be equal to zero

Question 24

Answer 24

Option C: 5,000 ohms

Question 25

Answer 25

Option A: 6

A charge of 1 Coulomb (C) is 6.28 x10^18. Hence 4+2 = 6 C

Question 26

Answer 26

Option C: working out the current flowing in the circuit by using ohms law and multiplying the result by the resistance

Question 27

Answer 27

Option B: 600

Q = IT where Q = charge in Coulombs
I = Current and T = time in seconds
Q = IT, Q = 5 * 120 = 600C

Question 28

Answer 28

Option B: 18A

Question 29

Answer 29

Option B: 600J

Work out power.
Current = 10V/10ohms = 1A (ohms law)
Power = Voltage * Current, Power = 10V * 1A , Power = 10 Watts

Energy = Power * Time where power is in watts and time is in seconds
Energy = 10 Watts * 60 seconds, Energy = 600 Joules