Question 14 Marks

Study the following electric circuit in which the resistors are arranged in three arms A, B and C:

i. Find the equivalent resistance of arm B .
ii. Calculate the equivalent resistance of the parallel combination of the arms B and C.
iii. Determine the current that flows through the ammeter.
OR
Determine the current that flows in the ammeter when the arm B is withdrawn from the circuit.

Answer

i. Since the resistance in arm $B$ are connected in series.
So, $\mathrm{R}_{\mathrm{B}}=5 \Omega+10 \Omega+15 \Omega$
$\mathrm{R}_{\mathrm{B}}=30 \Omega$
ii. Total resistance in arm C
$\mathrm{R}_{\mathrm{C}}=10 \Omega+20 \Omega+30 \Omega$
$\mathrm{R}_{\mathrm{C}}=60 \Omega$
Now as arm B and arm C are in parallel
Equivalent resistance $\frac{1}{R}=\frac{1}{R_{B}}+\frac{1}{R_{C}}$
$\frac{1}{R}=\frac{2+1}{60}=\frac{3}{60}$
$\mathrm{R}=20 \Omega$
iii. Total resistance in arm A
$\mathrm{R}_{\mathrm{A}}=5 \Omega+15 \Omega+20 \Omega$
$\mathrm{R}_{\mathrm{A}}=40 \Omega$
Now, Equivalent resistance of circuit
$\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{\mathrm{eq}}+\mathrm{R}$
$\mathrm{R}_{\text {eq }}=40+20=60 \Omega$
By ohm's law
$\mathrm{V}=\mathrm{IR}$
$I=\frac{V}{R}=\frac{6}{60}=0.1 \mathrm{~A}$
OR
If arm B is removed
Equivalent resistance in circuit
$\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{\mathrm{A}}+\mathrm{R}_{\mathrm{C}}$
$\mathrm{R}_{\text {eq }}=40+60=100 \Omega$
From ohm's law
$\mathrm{V}=\mathrm{IR} \\ I=\frac{V}{R}=\frac{6}{100}=0.06 \mathrm{~A}$

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