Answer to Question #294387 in Electric Circuits for Mwandium

Question

1. Determine the reading of the voltmeter shown in the figure. Consider the values of the
resistors as: R1 = 5Ω, R2 = 4.4Ω, R3 = 7.7Ω, R4 = 4.5Ω, R5 = 4.8Ω and that the source
provides a voltage V = 8.2V .
2. Let us be two cylindrical conductors connected in parallel, to which a potential
difference of V = 170V is applied. The two conductors are made of the same material,
but the first is 6 times the length of the second, and the radius of the second. The
resistance of the second is R2 = 469Ω. Determine the equivalent resistance.
3. In the circuit shown in the figure, the ideal ammeter measures I = 0.4A in the indicated
sense, and the ideal voltmeter measures a potential drop of V = 8.8V passing from b a a.
Determines the value of the emf ε2
.
Data: R1 = 56.2Ω, R2 = 23.3Ω, R3 = 27.4Ω.
1

Answer 1

When $R_1,R_2,R_3$

Series combination

$R_{eq}=R_1+R_2+R_3+R_4+R_5$

$$R_{eq}=5+4.4+7.7+4.5+4.8=26.4\Omega$$

$I=\frac{V}{R_{eq}}=\frac{8.2}{26.4}=0.31A$

$V_1=iR_1=5\times0.31=1.55V\\V_2=iR_2=4.4\times0.31=1.364V\\V_3=iR_3=7.7\times0.31=2.387V\\V_4=iR_4=0.31\times4.5=1.395V\\V_5=iR_5=0.31\times4.8=1.488V$

(2)

$R_1=\frac{\rho l_1}{A_1}$

$R_2=\frac{\rho l_2}{A_2}$

$\frac{R_2}{R_2}=\frac{l_2A_1}{l_1A_2}$

$\frac{R_2}{R_1}=\frac{6l_1\times 3.14\times36r_2^2}{l_1\times3.14\times r_2^2}=216\Omega$

$R_1=\frac{469}{216}$ =2.17$\Omega$

(3 )

$$e=V-iR\\e=8.8-0.4\times26.4=8.8-10.58=1.76V$$