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1.
The junction rule describest he conservation of which quantity? Note that this rule applies
only to circuits that are in steady state.
. current
voltage
resistance
The answer is current, as this rule says that the algebraic sum of the current is zero or
total inward current is equal to total outward current.
Part B
Apply the junction rule to the junction labeled with the number 1 (at the bottom of the
resistor of resistance
).
Answer in terms of given quantities, together with the meter readings
current
and
and the
.
- I1 + I2 + I3
I2 and I3 are incoming currents and I1 is outgoing current.
Part C
Apply the loop rule to loop 2 (the smaller loop on the right). Sum the voltage changes
across each circuit element around this loop going in the direction of the arrow.
Remember that the current meter is ideal.
Express the voltage drops in terms of
quantities.
I2R2 – I3R3
,
,
, the given resistances, and any other given
As the current in R2 is in the direction of the loop and that in R3 is opposite.
Part D
Now apply the loop rule to loop 1 (the larger loop spanning the entire circuit). Sum the
voltage changes across each circuit element around this loop going in the direction of the
arrow.
Express the voltage drops in terms of
quantities.
- Vb + I1R1 + I3R3
,
,
, the given resistances, and any other given
as the voltage gain at the cell and drop at the resistors.
2.
What is the effective resistance
of the two-resistor system?
Express the effective resistance in terms of
= R1 + R2
and
.
The two resistors are in series.
Three Resistors in Series
Now consider three resistors set up in series, as shown.
The resistances are
Part B
,
, and
Find the effective resistance,
, and the applied constant voltage is again
, of the three-resistor system.
.
Express the effective resistance in terms of
= R1 + R2 + R3
,
, and
.
3.
Which two of the following statements are true?
1. The voltage source provides a constant voltage, a part of which,
across resistor 1, and the remainder,
2. The voltage drops
, across resistor 2. Hence
across resistor 1 and
they are independent of the values of
voltage provided by the source. Hence
and
4. The currents
.
across resistor 2 are the same, and
. Both
and
are equal to the
.
3. The current provided by the voltage source splits: A part of it,
resistor 1, while the remainder,
, drops off
, flows through
, flows through resistor 2. Hence
through resistor 1 and
.
through resistor 2 are the same, and they
are equal to the current provided by the source. Hence
.
1 and 3
1 and 4
2 and 3
2 and 4
third answer is correct. Statements 2 and 3 are the correct as the two resistances are in
parallel.
4.
For the combination of resistors shown, find the equivalent resistance between points A
and B.
Express your answer in Ohms.
In the first diagram the resistances are in series and hence the equivalent resistance will
be the sum of the three.
R = 2 + 3 + 4 = 9 ohms.
=9
Part B
For the set-up shown, find the equivalent resistance between points A and B.
In the second diagram the two resistors are in parallel and the equivalent resistance R is
1
1
1
1 1 3


  
given by
R R1 R2 6 3 6
Or
R=2
Express your answer in Ohms.
=2
5.
Two resistors, with resistances of 25.0
and 21.0
are connected in parallel, and the
combination is connected across a DC line with a voltage of 240
What is the resistance of the parallel combination?
As above the resistance of the combination R is given by
1
1
1
1
1
46





R R1 R2 25 21 525
or
R = 525/46 =11.4 
11.4
Part B
What is the total current through the parallel combination?
The current will be I = V/R = 240/11.4 = 21.05 Amp.
21.0 A
Part C
What is the current through the first resistor?
I1 = V/R1 = 240/25 = 9.6 A
.Part A
9.6 A
Part D
What is the current through the second resistor?
I2 = V/R2 = 240/21 = 11.4 A
11.4 A
6.
The power rating of a resistor is the maximum power the resistor can safely dissipate
without too great a rise in temperature. The power rating of a 19.0
resistor is 5.00 .
What is the maximum allowable potential difference across the terminals of the resistor?
We know that the power P is given by
P = IV = I2R = V2/R
Hence V =
308.22 V
PR  5.00 *19 *103  308.22 V
Part B
A 9.70
resistor is to be connected across a potential difference of 120
rating is required?
. What power
P = V2/R = 120*120/(9.70*103) = 1.48 W
7.
In the circuit shown in the figure
.
, the voltage across the 2.00
resistor is 12.5
What is the emf of the battery?
Current in 2 ohm resistor is I = V/R = 12.5/2 = 6.25 A
The same current will be through 1 ohm and hence voltage across 1.00 ohm resistor will
be V1 = IR = 6.25*1.00 = 6.25 V
Hence total voltage drop across 1 and 2 ohm = 12.5 + 6.25 = 18.75 V
So the emf of the battery = 18.75 V
18.75 V
Part B
What is the current through the 6.00
resistor?
The same 18.75 V voltage will be dropped across 6.00 ohm resistor hence current
through it will be I2 = 18.75/6.00 = 3.125 A.
3.125 A
8.
How many 90-W, 120-V light bulbs can be connected to a 20-A, 120-V circuit without
tripping the circuit breaker? (Note: This description of a light bulb gives the power it
dissipates when connected to the stated potential difference; that is, a 25-W, 120-V light
bulb dissipates 25 W when connected to a 120-V line.)
The required voltage for the bulb is 120 V which is same as supply voltage and hence the
bulbs are to be connected in parallel.
The current through each bulb will be given by the relation
Or
P = IV
I = P/V = 90/120 = 0.75 A
Hence number of bulbs can be connected = 20/0.75 = 26.66
Hence maximum 26 bulbs can be connected in parallel.
26
9.
Three identical resistors are connected in series. When a certain potential difference is
applied across the combination, the total power dissipated is 26.0 .What power would
be dissipated if the three resistors were connected in parallel across the same potential
difference?
Let resistance of each resistor is R then the total resistance when they are in series will be
3R
The voltage is given by ( as above )
V  PRT  26.0 * 3R
now if the resistors are in parallel then the equivalent resistance will be R/3
hence power dissipated will be
P’ = V2/(R/3) =
234 W
10.
26 * 3R
 26 * 9  234 W
R/3
What happens to the charge on the capacitor immediately after the switch is thrown?
The electrons on the negative plate of the capacitor are held inside the capacitor by
the positive charge on the other plate.
Only the surface charge is held in the capacitor; the charge inside the metal plates
flows through the resistor.
The electrons on the negative plate immediately pass through the resistor and
neutralize the charge on the positive plate.
The electrons on the negative plate eventually pass through the resistor and
neutralize the charge on the positive plate.
Last statement is correct
The electrons on the negative plate eventually pass through the resistor and neutralize the
charge on the positive plate.
11. A capacitor with a capacitance of 454
65.2
is charged with charge of magnitude
on each plate. The capacitor is then connected to a voltmeter that has an internal
resistance of 1.23
.Part A
What is the current through the voltmeter just after the connection is made?
The voltage across the capacitor is given by
65.2 *10 9
 143.6V
454 *10 12
Hence the current at the time of contact will be
V = q/C =
I = V/R = 143.6/ (1.23*106) = 1.17*10-4 A
1.17*10-4 A
Part B
What is the time constant of this
circuit?
The time constant of RC circuit is given by
 = C*R = 454*10-12*65.2*10-9 = 2.96*10-17 second
2.96*10-17 s
12.
A capacitor with a capacitance of 4.30
that is initially uncharged is connected in
series with a resistor with a resistance of 6.50
and an EMF source with negligible
internal resistance which provides an EMF of 130
.
Part A
Just after the circuit is completed, what is the voltage drop across the capacitor?
As there is no charge on the capacitor voltage drop across it will be zero.
0V
Part B
Just after the circuit is completed, what is the voltage drop across the resistor?
The whole voltage is dropped across the resistor.
130 V
Part C
Just after the circuit is completed, what is the charge on the capacitor?
No charge
0C
Part D
Just after the circuit is completed, what is the current through the resistor?
Current through the resistor will be given by
V = IR
Or
I = V/R = 130/(6.50*103) = 0.02 A
0.02 A
An emf source with a magnitude of 120
capacitor with a capacitance of 5.80
, a resistor with a resistance of 99.0
are connected in series. As the capacitor charges,
when the current in the resistor is 0.850
plate of the capacitor?
, what is the magnitude of the charge on each
The voltage drop atcross the resistor V1 = 0.850*99.0 = 84.15 V
Hence voltage across capacitor V2 = 120 – 84.15 = 35.85 V
Hence the charge on the capacitor will be
Q = CV2 = 8.50*10-6*35.85 = 3.05*10-4 C
3.05*10-4 C
Thank God.
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