#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# The four arms of wheatstone bridge have the following resistanes,AB=100Ω, BC=10 Ω,CD=5 Ω, DA=60 Ω A Galvanometer of 15 Ω resistance is connected across BD. Calculate the current through the galvanometer when 10V is maintained across AC.

Grade:12

## 3 Answers

Saurabh Kumar
askIITians Faculty 2411 Points
5 years ago
.
yohanes
13 Points
2 years ago
1. The four arms of a Wheatstone bridge have the following resistances:
AB = 50Ω, BC = 5Ω, CD = 2Ω, DA = 25Ω. A galvanometer of 10Ω resistance is connected across
BD. If a potential difference of 20V is maintained across AC, using Thevenin’s theorem, calculate the current through the galvanometer and draw the Thevenin and Norton Equivalent circuit
ankit singh
askIITians Faculty 614 Points
one year ago
Considering the mesh ABDA, we have
100I1 + 15Ig – 60I2 = 0
or 20I1 + 3Ig – 12I2 = 0  ——(a)
Considering the mesh BCDB, we have
10 (I1 – Ig) – 5 (I2 + Ig) – 15Ig = 0
10I1 – 30Ig – 5I2 = 0
2I1 – 6Ig – I2 = 0 ——(b)
Considering the mesh ADCEA,
60I2 + 5 (I2 + Ig) = 10
65I2 + 5Ig = 10
13I2 + Ig = 2 ——(c)
Multiplying Eq. (b) by 10
20I1 – 60Ig – 10I2 = 0 ——(d)
From Eqs. (d) and (a) we have
63Ig – 2I2 = 0
I2 = 31.5Ig ——(e)
Substituting the value of I2 into Eq. [(c)], we get
13 (31.5Ig ) + Ig = 2
410.5 Ig = 2
Ig = 4.87 mA.

## ASK QUESTION

Get your questions answered by the expert for free