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Kirchhoff’s laws and Wheat stone’s bridge
- The magnitude and direction of the current in the circuit shown will be
(A)
3
7
A from a to b through e (B)
3
7
A from b to a through e
(C) 1A from b to a through e (D) 1A from a to b through e
Ans (D)
Since ( 10 ) ( 4 )
1 2
E VE V
So current in the circuit will be clockwise.
Applying Kirchoff's voltage law
1 i 10 4 2 i 3 i 0 i 1 A(atobviae)
Current ampere
R
V
- 0
6
10 4
- Kirchhoff's first law i. e. i 0 at a junction is based on the law of conservation of
(A) Charge (B) Energy (C) Momentum (D) Angular momentum
Ans (A)
- Kirchhoff's second law is based on the law of conservation of
(A) Charge (B) Energy (C) Momentum (D) Angular momentum
Ans (B)
- Two batteries of e.m.f. 4V and 8 V with internal resistances 1 and 2 are connected in a circuit with a resistance
of 9 as shown in figure. The current and potential difference between the points P and Q are
(A) A and 3 V
3
1
(B) A and 4 V
6
1
(C) A and 9 V
9
1
(D) A and 12 V
2
1
Ans (A)
Applying Kirchoff's voltage law in the given loop.
2 i 8 4 1 i 9 i 0
i A
3
1
Potential difference across PQ 9 3 V
3
1
- In the given circuit the current I 1
is
d c
a b
1
10 V
e
4 V
2
3
Q
2
1 4 V 8 V
P
9
r 1
r 2
1
3
2
10 V
i
4 V
E 2
E 1
e
a b
1
9
2
i
4 V
4 V 8 V
i
Q
P
(A) 0.4 A (B) – 0.4 A (C) 0.8 A (D) – 0.8 A
Ans (B)
The circuit can be simplified as follows
Applying KCL at junction A
3 1 2
i i i .….(i)
Applying Kirchoff’s voltage law for the loop ABCDA
30 40 40 0
1 3
i i
30 40 ( ) 40 0
1 1 2
i i i
7 4 4
1 2
i i .….(ii)
Applying Kirchoff’s voltage law for the loop ADEFA.
40 40 80 40 0
2 3
i i
40 40 ( ) 120
2 1 2
i i i
2 3
1 2
i i …….(iii)
On solving equation (ii) and (iii) i 0. 4 A
1
- In the balanced Wheatstone’s bridge circuit as shown in the figure, when the key is pressed, what will be the change
in the reading of the galvanometer?
(A) remains same (B) increased (C) decreased (D) none of these
Ans (A)
As the Wheatstone bridge is balanced, the pressing of key K makes no effect. The reading of the galvanometer G
remains the same
- The current I drawn from the 5 V source will be
(A) 0.67 A (B) 0.17 A (C) 0.33 A (D) 0.5 A
Ans (D)
30
40
40
40 V
I 3
I 1
I 2
80 V
i
i
i
i
A
B C
E
F
30
40
40 80 V
40 V
D
- The given figure shows a balanced Wheatstone’s network. Now, it is distributed by changing P to 11 . Which of
the following steps will not bring the bridge to balance again?
(A) increasing R by 2 (B) increasing Q by 10
(C) increasing S by 20 (D) making product RQ = 22200 ()
2
Ans (D)
When S is increased by 20 , we have
That is the bridge is not balanced
- Each of the resistances in the network shown in figure equals R. Find the resistance between two terminals A and
C
(A) 2R (B)
R
(C) R (D) R
2
Ans (C)
The network shown in figure is the equivalent network of the given network
It is a balanced Wheatstone bridge because
R R
R R
Hence the points B and D must be at the same potential. The resistance R in arm BD is ineffective
Total resistance along ADC = R + R = 2 R
Total resistance along ABC = R + R = 2 R
These two resistances form a parallel combination.
Effective resistance between A and C
2R 2R
R
2R 2R
- Six equal resistors, each of value R, are joined together as shown in figure. Calculate the equivalent resistance across
AB. If a supply of emf is connected across AB, compute the current through the arms AB.
(A)
0
R
(B)
R
(C)
0
R
(D)
R
Ans (D)
The equivalent resistance R across AB is given by
R 2R 2R R 2R R
or
R
R
Current through arm
AB
R R / 2 R
EMF, Internal Resistance, Terminal Potential Difference, Grouping of Cells, Comparison of EMF
of two cells, internal resistance, Electric Power, Meter Bridge, Potentiometer
- A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The
heat developed in it. The heat developed is doubled, if
(A) both the length and radius of wire are halved. (B) both length and radius of wire are doubled
(C) the radius of wire is doubled (D) the length of the wire is doubled
Ans (B)
2 2 2
2
V V V
P
R
A r
l l
or
2 2 2
V r r
P i.e., P
l l
When both l and r are doubled, P gets doubled
- A 25W-220 V bulb and a 100W-220V bulb are joined in series and connected to the mains. Which bulb will glow
brighter?
(A) 25 W bulb (B) 100 W bulb
(C) First 25 W bulb and then 100 W bulb (D) Both will glow with some brightness
Ans (A)
In the series circuit, same current flows through each bulb. But the 25 W bulb has a higher resistance
(R = V
2
/P). It produces more heat per second (P = I
2
R) and hence glows brighter than 100 W bulb
- An electric bulb is rated 220 V-100 W. The power consumed by it, when operated on 110 V, will be
(A) 25 W (B) 50 W (C) 75 W (D) 40 W
Ans (A)
2
V 220 220
R 484
P 100
When operated on 110 V
2
V 110 110
P 25 W
R 484
- A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be
(A) doubled (B) four times (C) one fourth (D) halved
Ans (A)
(A) 1 (B) 2 (C)
(D) 2.5
Ans (B)
In series,
1
2E
I
2 2r
In parallel,
2
E 2E
I
r 4 r
Since I 1 = I 2
2E 2E
4 r 2 2r
r 2
- When a resistance of 2 ohm is connected across the terminals of a cell, the current is 0.5 A. When the resistance is
increased to 5 ohm, the current is 0.25 A. The e.m.f. of the cell is
(A) 1.0 V (B) 1.5 V (C) 2.0 V (D) 2.5 V
Ans (B)
E
i ,
R r
we get
E
2 r
......(i)
E
5 r
.....(ii)
Dividing (i) by (ii), we get
5 r
2 r
r 1
E
E 1.5V
- A cell of internal resistance 1.5 and of e.m.f. 1.5 volt balances 500 cm on a potentiometer wire. If a wire of 15
is connected between the balance point and the cell, then the balance point will shift
(A) To zero (B) By 500 cm
(C) By 750 cm (D) None of the above
Ans (D)
Balance point has some fixed position on potentiometer wire. It is not affect by the addition of resistance between
balance point and cell.
- Two batteries, one of emf 18 volts and internal resistance 2 and the other of emf 12 volt and internal resistance
1 , are connected as shown. The voltmeter V will record a reading of s
(A) 15 volt (B) 30 volt (C) 14 volt (D) 18 volt
Ans (C)
Reading of voltmeter
1 2 2 1
eq
1 2
E r E r 18 1 12 2
E 14V
r r 1 2
- If the length of potentiometer wire is increased, then the length of the previously obtained balance point will
(A) Increase (B) Decrease
V
2
18 V
1
12 V
(C) Remain unchanged (D) Become two times
Ans (A)
When the length of potentiometer wire is increased, the potential gradient decreases and the length of previous
balance point is increased.
- Resistance in the two gaps of a meter bridge are 10 ohm and 30 ohm respectively. If the resistances are interchanged
the balance point shifts by
(A) 33.3 cm (B) 66.67cm (C) 25 cm (D) 50 cm
Ans (D)
100 l
S .R
l
Initially,
100 l
30 10 l 25cm
l
Finally,
100 l
10 30 l 75cm
l
So, shift = 50cm.
- A potentiometer wire is 10 m long and has a resistance of 18 . It is connected to a battery of emf 5 V and internal
resistance 2 . Calculate the potential gradient along the wire.
(A) 0.65 Vm
1
(B) 0.45 Vm
1
(C) 0.35 Vm
1
(D) 0.25 Vm
1
Ans (B)
Here l = 10 cm, R = 18 , = 5 V, r = 2
Current through the potentiometer wire,
I A
R r 18 2 20 4
Potential gradient
1
IR 1 18
0.45 Vm
l
- A potentiometer wire is supplied a constant voltage is 3 V. A cell of emf 1.08 V is balanced by the voltage drop
across 216 cm of the wire. Find the total length of the potentiometer wire
(A) 300 cm (B) 400 cm (C) 600 cm (D) 500 cm
Ans (C)
Here = 3 V, 1
= 1.08 V, l 1
= 216 cm, l =?
As
1
1 1 1
600 cm
l
l l
l
- The length of the potentiometer wire is 5 cm. It is connected to a battery of constant emf. For a given Leclanche
cell, the position of zero galvanometer deflection is obtained at 100 cm. If the length of the potentiometer wire be
made 8 m instead of 5 m, calculate the length of wire for zero deflection in the galvanometer for the same cell
(A) 1.6 m (B) 3.2 m (C) 2.8 m (D) 6.4 m
Ans (A)
Here l = 5 cm, l 1
= 100 cm = 1 m, l = 8 m, l 1
Let be the emf of the Leclanche cell
In first case,
1
IR
l
l
…(i)
In second case,
1
IR
f
l
l
…(ii)
- In a meter bridge, the balancing length from the left end (standard resistance of one ohm is in the right gap) is found
to be 20 cm. The value of the unknown resistance is
(A) 0. 8 (B) 0. 5 (C) 0. 4 (D) 0. 25
Ans (D)
80
20
1
X
4
1
X.
