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$15$
I) In the circuit shown, the cells have negligible internal resistance and the reading on the galvanometer is zero.
$\frac{6}{9}$
What is the value of resistor $R$ ?
A $2.0 \Omega$
B $6.0 \Omega$
C $12 \Omega$
D $18 \Omega$
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Answer from Sia

Posted 4 months ago

Solution

a

Given that the reading on the galvanometer is zero, the potential difference across the galvanometer must be zero. This implies that the potential at the point where the galvanometer is connected is the same for both loops

b

Apply Kirchhoff's Voltage Law (KVL) to the left loop:
$90V - 6\Omega \cdot I_1 - 9\Omega \cdot I_1 = 0$
$90V = 15\Omega \cdot I_1$
$I_1 = \frac{90V}{15\Omega} = 6A$

c

Apply Kirchhoff's Voltage Law (KVL) to the right loop:
$40V - 3\Omega \cdot I_2 - R \cdot I_2 = 0$
$40V = (3\Omega + R) \cdot I_2$

d

Since the potential at the galvanometer connection point is the same for both loops, the current $I_2$ in the right loop must be the same as $I_1$ in the left loop:
$I_2 = I_1 = 6A$

e

Substitute $I_2 = 6A$ into the right loop equation:
$40V = (3\Omega + R) \cdot 6A$
$40V = 18\Omega + 6A \cdot R$
$40V - 18\Omega = 6A \cdot R$
$22V = 6A \cdot R$
$R = \frac{22V}{6A} = 3.67\Omega$

f

The closest value to 3.67Ω among the given options is 6.0Ω

Answer

B

Key Concept

Kirchhoff's Voltage Law (KVL): The sum of the electrical potential differences (voltage) around any closed network is zero.

Explanation

By applying KVL to both loops and using the condition that the galvanometer reading is zero, we can determine the value of the unknown resistor $R$.

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