Question

Four resistors are connected to a battery as shown in the figure. The current in the battery is I the battery emf is E - 8.00 V, and the resistor values are R_1 = R, R_2 = 2R, R_3 = 4R, R_4 = 3R. Find the voltages across each resistor.

Answer 1

first find equivalent resistance

1)R2 &R3 are in series and result is parallel withR4 and again series with R1

R=(6R*3R)/9R=2R

2R+R=3R

I=V/Req=8/3R=2.66/R

VR1=2.66/R*(R)=2.66V

VR4=8V-2.66V=5.34V(as R1and R4 are parallel voltage across them is same)

VR3=5.33(4R/(2R+4R))=3.55V

VR2=5.33-3.55=1.78V

Answer 2

SOLUTION :

Equivalent of all resistors connected to the battery 

= R1 + (1 / (1/(R2 + R3) + 1/R4) 

= R + (1/(2R + 4R) + 1/3R)

= R + 1/( ( 1 + 2) / 6R) 

= R + 1/(1/2R)

= R + 2R

= 3R 

So, current flow from the battery, I 

= V / (3R) = 8.0 / (3R) 

= 2.667 / R  amps.

Voltage , v1, across resistor R1 

= I * R 

= 2.667 / R  * R 

= 2.667 volts .

So, voltage across resistors (R2 + R3)  

= 8.0 - 2.667 = 5.333 volts.

But, R2 + R3 = 2R + 4R = 6R

So, current through (R2 + R3) 

= 5.333 / (6R)

= 0.88883 / R

So, voltage v2 across R2 

= 0.88883 / R * 2R 

= 1. 778  volts 

So, voltage v3 across R3 

= 5.333 - 1.778 

= 3.555 volts 

Voltage v4 across R4 

= Voltage across (R2 + R3)

= 5.333 volts 

So, 

 v1 = 2.667 volts 

v2 = 1.778 volts 

v3 = 3.555 volts  

v4 = 5.333 volts 

(ANSWER).