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](https://tex.z-dn.net/?f=R_%7B1%7D%3D711%5C%20%5COmega)
![R_{2}=182\ \Omega](https://tex.z-dn.net/?f=R_%7B2%7D%3D182%5C%20%5COmega)
![R_{3}=663\ \Omega](https://tex.z-dn.net/?f=R_%7B3%7D%3D663%5C%20%5COmega)
![R_{4}=534\ \Omega](https://tex.z-dn.net/?f=R_%7B4%7D%3D534%5C%20%5COmega)
![R_{5}=265\ \Omega](https://tex.z-dn.net/?f=R_%7B5%7D%3D265%5C%20%5COmega)
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}}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7BR%7D%3D%5Cdfrac%7B1%7D%7BR_%7B2%7D%7D%2B%5Cdfrac%7B1%7D%7BR_%7B3%7D%7D%2B%5Cdfrac%7B1%7D%7BR_%7B4%7D%7D%2B%5Cdfrac%7B1%7D%7BR_%7B5%7D%7D)
Put the value into the formula
![\dfrac{1}{R}=\dfrac{1}{182}+\dfrac{1}{663}+\dfrac{1}{534}+\dfrac{1}{265}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7BR%7D%3D%5Cdfrac%7B1%7D%7B182%7D%2B%5Cdfrac%7B1%7D%7B663%7D%2B%5Cdfrac%7B1%7D%7B534%7D%2B%5Cdfrac%7B1%7D%7B265%7D)
![\dfrac{1}{R}=\dfrac{35501}{2806615}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7BR%7D%3D%5Cdfrac%7B35501%7D%7B2806615%7D)
![R=79.05\ \Omega](https://tex.z-dn.net/?f=R%3D79.05%5C%20%5COmega)
R and R₁ are connected in series
We need to calculate the equilibrium resistance
Using series formula
![R_{eq}=R_{1}+R](https://tex.z-dn.net/?f=R_%7Beq%7D%3DR_%7B1%7D%2BR)
![R_{eq}=711+79.05](https://tex.z-dn.net/?f=R_%7Beq%7D%3D711%2B79.05)
![R_{eq}=790.05\ \Omega](https://tex.z-dn.net/?f=R_%7Beq%7D%3D790.05%5C%20%5COmega)
We need to calculate the equivalent current
Using ohm's law
![i_{eq}=\dfrac{V}{R_{eq}}](https://tex.z-dn.net/?f=i_%7Beq%7D%3D%5Cdfrac%7BV%7D%7BR_%7Beq%7D%7D)
Put the value into the formula
![i_{eq}=\dfrac{64.5}{790.05}](https://tex.z-dn.net/?f=i_%7Beq%7D%3D%5Cdfrac%7B64.5%7D%7B790.05%7D)
![i_{eq}=0.0816\ A](https://tex.z-dn.net/?f=i_%7Beq%7D%3D0.0816%5C%20A)
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}](https://tex.z-dn.net/?f=i_%7Beq%7D)
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](https://tex.z-dn.net/?f=V%3DI_%7Beq%7D%5CtimesR)
Put the value into the formula
![V=0.0816\times79.05](https://tex.z-dn.net/?f=V%3D0.0816%5Ctimes79.05)
![V=6.45\ Volt](https://tex.z-dn.net/?f=V%3D6.45%5C%20Volt)
In resistors parallel combination voltage distribution in each part will be same.
So, ![V_{2}=V_{3}=V_{4}=V_{5}=6.45 V](https://tex.z-dn.net/?f=V_%7B2%7D%3DV_%7B3%7D%3DV_%7B4%7D%3DV_%7B5%7D%3D6.45%20V)
We need to calculate the current at H point
Using ohm's law
![i_{h}=\dfrac{V_{5}}{R_{5}}](https://tex.z-dn.net/?f=i_%7Bh%7D%3D%5Cdfrac%7BV_%7B5%7D%7D%7BR_%7B5%7D%7D)
Put the value into the formula
![i_{h}=\dfrac{6.45}{265}](https://tex.z-dn.net/?f=i_%7Bh%7D%3D%5Cdfrac%7B6.45%7D%7B265%7D)
![i_{h}=0.0243\ A](https://tex.z-dn.net/?f=i_%7Bh%7D%3D0.0243%5C%20A)
Hence, The current in R₁ is 0.0816 A.
The current at H point is 0.0243 A.