Physics Test on Electricity : Wheatstone Bridge

Study the Wheatstone Bridge below to answer the following questions.

1. In the balanced Wheatstone bridge shown, what is the current flowing through the galvanometer (G)?

0 A
0.4 A
0.8 A
1.2 A

2. What is the total resistance of the left branch (A–B–C), consisting of resistors R₁ and R₃ connected in series?

20 Ω
25 Ω
30 Ω
35 Ω

3. What is the voltage drop across resistor R₁?

4 V
6 V
8 V
10 V

4. What is the power dissipated by resistor R₂?

1.6 W
3.2 W
6.4 W
8.0 W

5. If resistor R₃ operates continuously for 5 minutes, how much electrical energy does it consume?

1440 J
2160 J
2880 J
3600 J

📖 View Solutions here

A Wheatstone bridge is an electrical circuit arranged in a diamond shape of four resistors, used to measure an unknown resistance by adjusting one of the resistors until the electric current flowing through a central meter (galvanometer) becomes exactly zero. This "balanced" condition means the two mid-points of the circuit are at the same voltage, allowing the unknown resistance to be calculated from the known ratios of the other three resistors. In simpler terms: It is a precision measuring tool that finds an unknown resistance by comparing it to known resistances, using the absence of current (a "null" reading) as the sign that the measurement is correct.

Detailed Solution - Question 1

Question: In the balanced Wheatstone bridge shown, what is the current flowing through the galvanometer (G)?

Step 1: Check whether the bridge is balanced.

$$ \frac{R_1}{R_2}=\frac{10}{20}=0.5 $$

$$ \frac{R_3}{R_4}=\frac{15}{30}=0.5 $$

Since

$$ \frac{R_1}{R_2}=\frac{R_3}{R_4}, $$

the Wheatstone Bridge is balanced.

Step 2: Determine the galvanometer current.

In a balanced Wheatstone bridge, the potential difference across the galvanometer is zero.

$$ V_B=V_D $$

Therefore,

$$ V_G=0\text{ V} $$

Using Ohm's Law,

$$ I=\frac{V}{R} $$

$$ I_G=\frac{0}{R_G}=0\text{ A} $$

Answer: A. 0 A

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Detailed Solution - Question 2

Question: What is the total resistance of the left branch (A–B–C), consisting of resistors R₁ and R₃ connected in series?

Step 1: Add the series resistances.

$$ R_{Left}=R_1+R_3 $$

Substitute the values.

$$ R_{Left}=10+15 $$

$$ R_{Left}=25\,\Omega $$

Answer: B. 25 Ω

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Detailed Solution - Question 3

Question: What is the voltage drop across resistor R₁?

Step 1: Calculate the current in the left branch.

Total resistance of the left branch:

$$ R=10+15=25\,\Omega $$

Using Ohm's Law,

$$ I=\frac{V}{R} $$

$$ I=\frac{20}{25} $$

$$ I=0.8\text{ A} $$

Step 2: Calculate the voltage across R₁.

$$ V=IR $$

$$ V=0.8\times10 $$

$$ V=8\text{ V} $$

Answer: C. 8 V

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Detailed Solution - Question 4

Question: What is the power dissipated by resistor R₂?

Step 1: Calculate the current in the right branch.

Total resistance:

$$ R=20+30=50\,\Omega $$

$$ I=\frac{20}{50} $$

$$ I=0.4\text{ A} $$

Step 2: Calculate the power.

Use the power formula:

$$ P=I^2R $$

$$ P=(0.4)^2(20) $$

$$ P=0.16\times20 $$

$$ P=3.2\text{ W} $$

Answer: B. 3.2 W

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Detailed Solution - Question 5

Question: If resistor R₃ operates continuously for 5 minutes, how much electrical energy does it consume?

Step 1: Calculate the current through R₃.

Since R₁ and R₃ are connected in series, the current is

$$ I=0.8\text{ A} $$

Step 2: Calculate the power dissipated by R₃.

$$ P=I^2R $$

$$ P=(0.8)^2(15) $$

$$ P=0.64\times15 $$

$$ P=9.6\text{ W} $$

Step 3: Convert the time into seconds.

$$ t=5\times60=300\text{ s} $$

Step 4: Calculate the electrical energy.

$$ E=Pt $$

$$ E=9.6\times300 $$

$$ E=2880\text{ J} $$

Answer: C. 2880 J