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
[tex]R_{1}=711\ \Omega[/tex]
[tex]R_{2}=182\ \Omega[/tex]
[tex]R_{3}=663\ \Omega[/tex]
[tex]R_{4}=534\ \Omega[/tex]
[tex]R_{5}=265\ \Omega[/tex]
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
[tex]\dfrac{1}{R}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\dfrac{1}{R_{4}}+\dfrac{1}{R_{5}}[/tex]
Put the value into the formula
[tex]\dfrac{1}{R}=\dfrac{1}{182}+\dfrac{1}{663}+\dfrac{1}{534}+\dfrac{1}{265}[/tex]
[tex]\dfrac{1}{R}=\dfrac{35501}{2806615}[/tex]
[tex]R=79.05\ \Omega[/tex]
R and R₁ are connected in series
We need to calculate the equilibrium resistance
Using series formula
[tex]R_{eq}=R_{1}+R[/tex]
[tex]R_{eq}=711+79.05[/tex]
[tex]R_{eq}=790.05\ \Omega[/tex]
We need to calculate the equivalent current
Using ohm's law
[tex]i_{eq}=\dfrac{V}{R_{eq}}[/tex]
Put the value into the formula
[tex]i_{eq}=\dfrac{64.5}{790.05}[/tex]
[tex]i_{eq}=0.0816\ A[/tex]
We know that,
In series combination current distribution in each resistor will be same.
So, Current in R and R₁ will be equal to [tex]i_{eq}[/tex].
The current at h point will be equal to current in R₅
We need to calculate the voltage in R
Using ohm's law
[tex]V=I_{eq}\timesR[/tex]
Put the value into the formula
[tex]V=0.0816\times79.05[/tex]
[tex]V=6.45\ Volt[/tex]
In resistors parallel combination voltage distribution in each part will be same.
So, [tex]V_{2}=V_{3}=V_{4}=V_{5}=6.45 V[/tex]
We need to calculate the current at H point
Using ohm's law
[tex]i_{h}=\dfrac{V_{5}}{R_{5}}[/tex]
Put the value into the formula
[tex]i_{h}=\dfrac{6.45}{265}[/tex]
[tex]i_{h}=0.0243\ A[/tex]
Hence, The current in R₁ is 0.0816 A.
The current at H point is 0.0243 A.
