Answer to Question #158938 in Electric Circuits for Arslan habib

Question

For the circuit shown in fig. Both voltmeter and ammeter are idealized, the battery has no appreciable internal resistance, and the ammeter reads 5 A.

a. What does the voltmeter V read?.

b. What is the total current I of the circuit?.

c. What is the total voltage of the battery?

d. What is the total resistance R of the circuit?

25.022

A

15.02

MW

45.012

15.022

10.022

35.022 V = ?

Answer 1

d) First of all we must find total resistance:

$R_1 = 25\varOmega \space R_2 = 15\varOmega \space R_3 = 15\varOmega \space R_4 = 10\varOmega \space R_5 = 45\varOmega \space R_6 = 35\varOmega \space$

The $R_3 \space and \space R_4$ resistors are connected in series $R_{3,4} = R_3+R_4 = 15+10=25\varOmega$

The $R_1, R_2 \space and \space R_{3,4}$ resistors are connected parallel.

$\large\frac{1}{R_{1,2,3,4}}= \frac{1}{R_1} +\frac{1}{R_2}+\frac{1}{R_{3,4}} = \frac{1}{25}+\frac{1}{15}+\frac{1}{25}= \frac{11}{75} \to R_{1,2,3,4} = \frac{75}{11}\varOmega$

The $R_5 \space and \space R_{1,2,3,4}$ are connected in series $R_{1,2,3,4,5} = R_{1,2,3,4}+R_5 = 45+\frac{75}{11} = 51\frac{9}{11}\varOmega$

$\large\frac{1}{R_t}= \frac{1}{R_{1,2,3,4,5}}+\frac{1}{R_6} = \frac{11}{560}+\frac{1}{35}=\frac{27}{560} \to$ $R_t = 20\frac{20}{27}\varOmega$

a) we have $I_1 = 5A$       $U_1 = I_1*R_1 = 125V$

$U_{1,2,3,4} = U_1 = U_2=U_{3,4}=125V$  because of they are connected parallel

$I_{1,2,3,4} = \large\frac{U_{1,2,3,4}}{R_{1,2,3,4}}=\frac{125*11}{75}=\frac{55}{3}$

$I_{1,2,3,4} = I_5=I_{1,2,3,4,5}=\frac{55}{3}$ because of they are connected in series

$U_5 = I_5*R_5=\frac{55}{3}*45=825V$

b) $U_{1,2,3,4,5} = U_{1,2,3,4}+U_{5}=825+125=1000V$

$U_t =U_{1,2,3,4,5} = U_6 = 1000V$ because of they are connected parallel

c) $I_t = \large\frac{U_t}{T_t} = \frac{1000*27}{560} =$ 48.2A