9810252244
Wheatstone bridge (theory+numericals)
Metre bridge (theory)
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Wheatstone bridge: It is an arrangement of four resistances used to determine one of
these resistances quickly and accurately in terms of the remaining three resistances. A
Wheatstone bridge consists of four resistances P, Q ,R and S ; connected to form the
arms of a quadrilateral ABCD. A battery of emf e is connected between points A and C
and a sensitive galvanometer between B and D, as shown in Fig.
The resistances are so adjusted that no current flows through the galvanometer. The
bridge is then said to be balanced. In the balanced condition,
P R
Q S
This is Wheatstone bridge principle.
Wheatstone bridge: It is an arrangement of four resistances used to determine one of
these resistances quickly and accurately in terms of the remaining three resistances. A
Wheatstone bridge consists of four resistances P, Q ,R and S ; connected to form the
arms of a quadrilateral ABCD. A battery of emf e is connected between points A and C
and a sensitive galvanometer between B and D, as shown in Fig.
The resistances are so adjusted that no current flows through the galvanometer. The
bridge is then said to be balanced. In the balanced condition,
P R
Q S
This is Wheatstone bridge principle.
Derivation of balance condition from Kirchhoff's laws: In accordance with
Kirchhoff’s first law, the currents through various
branches are as shown in Fig.
Applying Kirchhoff's second law to the loop ABDA,
we get
– I1 P – Ig G + I 2 R = 0
where G is the resistance of the galvanometer.
Again applying Kirchhoff's second law to the loop
BCDB, we get
– ( I1 – Ig ) Q + (I2 + Ig) S + GIg = 0
In the balanced condition of the bridge, Ig = 0 . The
above equations become
2
– I1P + I2R =0
or I1 P = I2R ….(i)
and –I1 Q + I 2 S = 0
or I1Q = I2 S ...(ii)
On dividing equation (i) by (ii), we get
I1P I 2 R P R
;
I1Q I 2 S Q S
This proves the condition for the balanced Wheatstone bridge. For the galvanometer to
give zero deflection.
Sensitivity of a Wheatstone bridge: A Wheatstone bridge is said to be sensitive if it
shows a large deflection in the galvanometer for a small change of resistance in the
resistance arm.
The sensitivity of the Wheatstone bridge depends on two factors :
(i) Relative magnitudes of the resistances in the four arms of the bridge. The bridge
is most sensitive when all the four resistances are of the same order.
(ii) Relative positions of battery and galvanometer.
Advantages of a Wheatstone bridge method of measuring resistance over other
methods.
(i) It is a null method, hence the result is free from the effect of extra resistances (cell
resistances) of the circuit.
(ii) Being null method, it is easier to detect a small change in deflection than to read a
deflection directly.
[1] What is the resistance between points A and B in the circuit shown in Fig. ?
SOL: The network represents a balanced Wheatstone bridge and hence the resistance
of 5Ω connected across the, diagonal points can be considered as an open path. As
such, the network is equivalent to the circuit drawn in Fig.
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The network is now a parallel combination of two resistive paths. The resistance of
each resistive path is 5+ 5 i.e. 10 Ω.Hence, the net resistance between the points A and
B,
[2] Six equal resistors, each of 5 Ω are connected to a
battery of e.m.f. 3 V as shown in Fig. What is the current
drawn by the circuit from
the battery ?
SOL: The given network can
be arranged as shown in Fig.
It is a balanced Wheatstone
bridge. effective resistance of
the network, R = 2.5 Ω
Therefore, current drawn by the circuit,
3
I 1.2 A
2.5
[3] Calculate the equivalent resistance between A and B of the network shown in
Fig.
SOL: The upper five resistances between A and B will form a
balanced Wheatstone bridge. Hence there will be no current
in arm CO. The equivalent circuit will be as shown in Fig.
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R
RAB
2
[4] For the network shown in the Fig. , determine the value of R and the current
through it, if the current through the branch AO is zero.
SOL: The given network can be rearranged as shown in Fig. As the current through the
branch AO is zero, the bridge is balanced i.e.
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[5] The galvanometer, in each of the two given circuits, does not show any
deflection. Find the ratio of the resistors R1 and R2, used in these two
circuits.[CBSE Sample Paper 2013][All India 2013]
SOL: For circuit 1, we have, (from the Wheatstone bridge balance condition),
In circuit 2, the interchange of the positions of the battery and the galvanometer, does
not change the (wheatstone Bridge) balance condition.
[6] The given Wheatstone bridge is showing no deflection in the galvanometer
joined between the points B and D Fig. Calculate the value of R.
6
SOL: Since the galvanometer shows no deflection therefore, Wheatstone bridge is
balanced. Applying condition for balanced Wheatstone bridge,
[7] Six resistors, each of value 6 , are joined
together in a circuit, as shown in Fig. Calculate
equivalent resistance across the points A and B .
If a cell of emf 3V is connected across AB ,
compute the current through the arms AB and DF
of the circuit [CBSE 2003]
Ans: 3 , 0.5 A, zero.
[8]
Find the value of the unknown resistance X, in the
following circuit, if no current flows through the
section AO, Fig. Also calculate the current drawn by
the circuit from the battery of emf 6 V and
negligible internal resistance.
[CBSE 2002]
SOL: When no current flows through the arm AO, the
Wheatstone bridge is balanced. Then
[9] An electrical circuit is shown in the figure. Calculate the potential difference
across the resistor of 400 Ω , as will be measured by the voltmeter V of resistance
400 Ω .
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SOL:
This current is divided equally between the arms AB , BC and AD , DC (Fig. c). Thus
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Metre bridge or slide wire bridge:
It is the simplest practica1 application of the Wheatstone bridge that is used to
measure an unknown resistance.
Principle Its working is based on the principle of Wheatstone bridge. When the
bridge is balanced,
P R
Q S.
Construction: It consists of usually one metre long magnanin wire of uniform
cross-section, stretched along a metre scale fixed over a wooden board and with
its two ends soldered to two L-shaped thick copper strips A and C. Between these
two copper strips, another copper strip is fixed so as to provide two gaps ab and
a1 b1 . A resistance box R.B. is connected in the gap ab and the
unknown resistance S is connected in the gap a 1 b 1 . A source of emf E is
connected across AC. A
movable jockey and a galvanometer are connected across BD, as shown in Fig.
Working: After taking out a suitable resistance R from the resistance box, the
jockey is moved along the wire AC till there is no deflection in the galvanometer.
This is the balanced condition of the Wheatstone bridge. If P and Q are the
resistances of the parts AB and BC of the wire, then for the balanced condition of
the bridge, we have
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P R
Q S
Let total length of wire AC =100 cm and AB= l cm, then BC =(100 – l ) cm.
Since the bridge wire is of uniform cross-section, therefore,
resistance of wire length of wire
P resistance of AB l l
or Q resistance of BC (100 l ) 100 l
where is the resistance per unit length of the wire.
R l R(100 l )
S
Hence S 100 l or l
Knowing l and R, unknown resistance S can be determined.
Determination of resistivity. If r is the radius of the wire and l ' its length, then
resistivity of
its material will be
SA S r 2
l' l'
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