Question

The ratio of the amounts of heat developed in the four arms of a balance Wheatstone bridge, when the arms have resistances \(P = 100\,\Omega\), \(Q = 10\,\Omega\), \(R = 300\,\Omega\) and \(S = 30\,\Omega\) respectively is

• A. \(3:30:1:10\)
• B. \(30:3:10:1\)
• C. \(30:10:1:3\)
• D. \(30:3:1:10\)

Hint:

In a balanced Wheatstone bridge, no current flows through the galvanometer, so currents split only in two branches and heat ratio is found using \(H \propto I^2R\).

Answer 1

The Correct Option is B

Step 1: Condition of balanced Wheatstone bridge.
A Wheatstone bridge is balanced when:
\[ \frac{P}{Q} = \frac{R}{S} \] Here,
\[ \frac{100}{10} = 10,\quad \frac{300}{30} = 10 \] So the bridge is balanced.
Step 2: Current distribution in balanced bridge.
In balanced condition, no current flows through the galvanometer.
Thus, the current \(I\) divides into two branches:
- One branch through \(P\) and \(Q\)
- Second branch through \(R\) and \(S\)
Let current through branch \(P,Q\) be \(I_1\) and through \(R,S\) be \(I_2\).
Step 3: Heat developed in each resistor.
Heat developed \(H \propto I^2Rt\).
Since time \(t\) is same for all arms, ratio depends on \(I^2R\).
Step 4: Find branch currents using equivalent resistance.
Resistance of branch \(P,Q\):
\[ P+Q = 100+10 = 110\,\Omega \] Resistance of branch \(R,S\):
\[ R+S = 300+30 = 330\,\Omega \] So current ratio:
\[ \frac{I_1}{I_2} = \frac{330}{110} = 3 \Rightarrow I_1 = 3I_2 \] Step 5: Heat ratios in each resistor.
Now,
\[ H_P \propto I_1^2 P,\quad H_Q \propto I_1^2 Q \] \[ H_R \propto I_2^2 R,\quad H_S \propto I_2^2 S \] Substitute \(I_1 = 3I_2\):
\[ H_P : H_Q : H_R : H_S = (9I_2^2)(100) : (9I_2^2)(10) : (I_2^2)(300) : (I_2^2)(30) \] \[ = 900 : 90 : 300 : 30 \] Divide all by 30:
\[ = 30 : 3 : 10 : 1 \] Final Answer: \[ \boxed{30:3:10:1} \]