In the circuit shown in the figure, find the total resistance of the c

1.	In the circuit shown in the figure, find the total resistance of the circuit and the current in the arm CD.
Answer» It can be seen that resistances BC and CD are in series and their combination is in parallel with AD. Then \(\frac{1}{R_p}=\frac{1}{6}+\frac{1}{3}\) \(\Rightarrow\) R_p = \(2\Omega\) Total resistance of circuit is 2 + 3 = 5Ω (Due to capacitor, resistor 3Ω in EF will not be counted) Total current = \(\frac{15}{5}=3A.\) This current gets divided at junction A. Voltage across DF = 3 Ω × 3 A = 9 V and Voltage across AD = 15 – 9 = 6 V I across CD = \(\frac{6\,V}{3\,+\,3}=1\,A\) Hence, current through arm CD = 1 A.

In the circuit shown in the figure, find the total resistance of the circuit and the current in the arm CD.

Answer»

It can be seen that resistances BC and CD are in series and their combination is in parallel with AD.

Then \(\frac{1}{R_p}=\frac{1}{6}+\frac{1}{3}\)

\(\Rightarrow\) R_p = \(2\Omega\)

Total resistance of circuit is 2 + 3 = 5Ω

(Due to capacitor, resistor 3Ω in EF will not be counted)

Total current = \(\frac{15}{5}=3A.\)

This current gets divided at junction A.

Voltage across DF = 3 Ω × 3 A = 9 V and Voltage across AD = 15 – 9 = 6 V

I across CD = \(\frac{6\,V}{3\,+\,3}=1\,A\)

Hence, current through arm CD = 1 A.