Wheatstone Bridge and Meter Bridge Numerical Class-12 Nootan ISC Physics Ch-6 DC Circuits and Measurements. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Wheatstone Bridge and Meter Bridge Numerical Class-12 Nootan ISC Physics Ch-6 DC Circuits and Measurements
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-6 | DC Circuits and Measurements. |
| Topics | Numericals on Wheatstone Bridge and Meter Bridge |
| Academic Session | 2025-2026 |
Numericals on Wheatstone Bridge and Meter Bridge
Class-12 Nootan ISC Physics Ch-6 DC Circuits and Measurements
Que-26. Calculate the current drawn from the battery in the given network.
Ans-26 The equivalent circuit is shown in the figure.
The five resistors form a balanced Wheatstone’s bridge
R1/R5 = R4/R3
So, R2 is ineffective.
The effective resistance of R1 and R5 in series,
R′= R1+R5 = 1+2 = 3Ω
The effective resistance of R4 and R3 in series is
R′′ = R4+R3 = 2+4 = 6Ω
Therefore, equivalent resistance of network between A and B is
RAB = (R′R′′)/(R′+R′′) = (3×6)/(3+6) = 2Ω
Current drawn from battery,
I = E/RAB = 4/2 = 2A.
Que-27. The given circuit diagram represents balanced Wheatstone’s bridge, calculate R.
Ans-27 resistance of parallel combination = X
then according to rule of wheat stone bridge
6/X = 3/1.5 ⇒ X = 3Ω
again X is made of parallel combination of R and 15Ω
X = (Rx15)/(R+15) = 3
15R = 3R+45
15R-3R = 45
12R = 45
R = 45/12 = 3.75Ω.
Que-28. Calculate the currents I, I1 and I2 in the given Wheatstone’s network of resistors. The internal resistance of the 6 V battery is 0.4 Ω
Ans-28 It is a balanced Wheat Stone Bridge
net resistance R = (6×9)/(6+9) = 54/15 = 3.6Ω
i = E/(R+r) = {6/(3.6+0.4)} = 6/4 = 1.5 A.
I1/I2 = 9/6 = 3/2
I1 = 3/5 x 1.5 = 0.9 A.
I2 = 1.5 – 0.9 = 0.6 A.
Que-29. The adjoining Wheatstone’s and the point C is earthed. (a) Find the potentials at points B and D. (b) If galvanometer a be connected across BD, what will be the direction of current in it? (c) For what value of the resistor DC would the bridge be balanced?
Ans-29 current through arm ABC = 60/30 = 2A
current through arm ACD = 60/10 = 6A
(a) VA – Vb =iR
60- VB =2×10
VB = 60-20 = 40V
and VA – VD = 3 x 6
60- VD = 18
VD = 60 – 18 = 42V
(b) since D is at higher potential therefore current will pass from D to B
(c) let for balance bridge R(DC)is X then by the rule of balanced wheatstone bridge
10/20 = 3/X
i.e. X = 6 Ω.
Que-30. In determining the resistance of a given wire with the help of metre bridge, the null-point is obtained distance of 40 cm on taking out 2 ohm resistance from the resistance box. What is the resistance of the wire? To obtain the null-point at a distance of 60 cm, how much resistance in ohm, will have to be taken out from the resistance box?
Ans-30 According to rule of metre bridge
2/R = 40/60
R = (2×60)/40 = 3Ω
To have null point at 60cm
Let x resistance should be taken out
X/3 = 60/40
X = (6×3)/4 = 4.5Ω
— : End of Wheatstone Bridge and Meter Bridge Numerical Class-12 Nootan ISC Physics Ch-6 DC Circuits and Measurements. :–
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