Question

The circuit to the right consists of a battery ( V 0 = 64.5 V) (V0=64.5 V) and five resistors ( R 1 = 711 (R1=711 Ω, R 2 = 182 R2=182 Ω, R 3 = 663 R3=663 Ω, R 4 = 534 R4=534 Ω, and R 5 = 265 R5=265 Ω). Find the current passing through each of the specified points. -g

Answer 1

Answer:

The current in R₁ is 0.0816 A.

The current at H point is 0.0243 A.

Explanation:

Given that,

Voltage = 64.5

Resistance is

$R_{1}=711\ \Omega$

$R_{2}=182\ \Omega$

$R_{3}=663\ \Omega$

$R_{4}=534\ \Omega$

$R_{5}=265\ \Omega$

Suppose, the specified points are R₁ and H.

According to figure,

R₂,R₃,R₄ and R₅ are connected in parallel

We need to calculate the resistance

Using parallel formula

$\dfrac{1}{R}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\dfrac{1}{R_{4}}+\dfrac{1}{R_{5}}$

Put the value into the formula

$\dfrac{1}{R}=\dfrac{1}{182}+\dfrac{1}{663}+\dfrac{1}{534}+\dfrac{1}{265}$

$\dfrac{1}{R}=\dfrac{35501}{2806615}$

$R=79.05\ \Omega$

R and R₁ are connected in series

We need to calculate the equilibrium resistance

Using series formula

$R_{eq}=R_{1}+R$

$R_{eq}=711+79.05$

$R_{eq}=790.05\ \Omega$

We need to calculate the equivalent current

Using ohm's law

$i_{eq}=\dfrac{V}{R_{eq}}$

Put the value into the formula

$i_{eq}=\dfrac{64.5}{790.05}$

$i_{eq}=0.0816\ A$

We know that,

In series combination current distribution in each resistor will be same.

So, Current in R and R₁ will be equal to $i_{eq}$

The current at h point will be equal to current in R₅

We need to calculate the voltage in R

Using ohm's law

$V=I_{eq}\timesR$

Put the value into the formula

$V=0.0816\times79.05$

$V=6.45\ Volt$

In resistors parallel combination voltage distribution in each part will be same.

So, $V_{2}=V_{3}=V_{4}=V_{5}=6.45 V$

We need to calculate the current at H point

Using ohm's law

$i_{h}=\dfrac{V_{5}}{R_{5}}$

Put the value into the formula

$i_{h}=\dfrac{6.45}{265}$

$i_{h}=0.0243\ A$

Hence, The current in R₁ is 0.0816 A.

The current at H point is 0.0243 A.