Download 10 Ω

KEY
Name: _______________________
Electric Circuit Homework
1. Draw a circuit diagram of two 20 Ohm resistors in series with each other powered by a 10 volt
battery. Place an ammeter in the circuit and a voltmeter across one of the resistors.
20 Ω
20 Ω
10 V
V
A
2. Calculate the total resistance of the circuit and the reading of the ammeter.
RT = R1 + R2
RT = 20 + 20 = 40 Ω
I=V/R
I = 10 / 40 = 0.25 A
3. Find the power consumed by each resistor and the total power delivered by the battery.
V = IR = (0.25) (20) = 5 V
P = VI
P = (5) (0.25) = 1.25 W
Total Power:
P = VI = (10) (0.25) = 2.5 W
or
P = I2R
P = (0.25)2 (20) = 1.25 W
or
P = P1 + P2 = 1.25 + 1.25 = 2.5 W
4. In the circuit shown on the right, the current through each ammeter is zero
before the switch is thrown. What is the current through each ammeter
after the switch is thrown and what is the total power delivered by the
battery. (hint, find the total resistance to compute the total current)
1/RT = 1/R1 + 1/R2 = 0.10 + 0.10 = 0.20
RT = 5 Ω
IT = VT / RT = 6 / 5 = 1.2 A
I1 = V1 / R1 = 6 / 10 = 0.60 A
5. What is the voltage across the lower 10 Ohm resistor? Across the upper 10 Ohm resistor?
Since resistors are in parallel with the battery, potential difference across each is 6 V.
6. The circuit to the right consists of a pair of 20 Ohm
resistors in series with a pair of 10 Ohm resistors in
series with a 30 Ohm resistor. What is the total
(equivalent) resistance of the circuit?
Parallel Resistors:
1/20 + 1/20 = 2/20  Flip back  20/2 = 10 Ω
1/10 + 1/10 = 2/10  Flip back  10/2 = 5 Ω
Series:
RT = R1 + R2
RT = 30 + 5 + 10 = 45 Ω
7. A 5 Ohm resistor is in parallel with a 10 Ohm resistor, find the total equivalent resistance.
Flip it

Add it

Flip it back 
1/5 and 1/10
1/5 + 1/10 = 3/10
10/3 = 3.3 Ω
5Ω
10 Ω
8. The circuit below contains three identical light bulbs. When the switch is closed, which of the
following will be true?
Bulb 1 will burn
a. brighter than before.
b. the same as before.
c. less bright than before.
Bulb 2 will burn
a. brighter than before.
b. the same as before.
c. less bright than before.
Bulbs 2 and 3 are in parallel with each other, so their combined resistance is less than
one bulb alone. Since the total resistance of the circuit is lower, the total current is
higher, bulb 1 will burn brighter. Bulbs 2 and 3 will each only have half the current that
goes through Bulb 1, so they will both be dimmer than Bulbs 1 and 2 were before closing
the switch.