Question

If the circuit shown in Figure 1 is equivalent to that shown in Figure 2, then,

• A. the emf V is 5 V
• B. the resistance R is about 3 $\Omega$
• C. the current through 5 $\Omega$ is about $\frac{2}{3}$ A
• D. the rate of dissipation of heat in R is about 1.1 Js${}^{-1}$

Answer 1

Answer: (A), (C), (D)

the rate of dissipation of heat in R is about 1.1 Js${}^{-1}$
the current through 5 $\Omega$ is about $\frac{2}{3}$ A
the emf V is 5 V

(i)Consider the given circuit as shown in figure.

applying Kirchhoff's law,

In loop abAca, $3I_1=21$

$\therefore I_1=7A$

In loop abBca, $7I_2=21$

$\therefore I_2=3A$

Now,

for branch AcB

$V_A-2I_1+3I_2=V_B$

$\therefore V_A-V_B=2I_1-3I_2=14-9=5V$

(ii) Resistance R is the resistance between A and B when the source of emf 21 V is shorted.

$\therefore R=\frac{50}{21}\Omega$

(iii) Current through the branch AB is

$I=\frac{5}{(5+\frac{50}{21})}=\frac{1}{(1+\frac{10}{21})}$

$\therefore I\simeq \frac{2}{3}A$

(iv) Rate of heat dissipation is

$H=\frac{4}{9}\times \frac{50}{21}\simeq \frac{10}{9}=1.1J{s}^{-1}$