A 6volt battery of internal resistance 0.5ohm is connected to three resistors of value, 3ohm, 4ohm, and 5ohm. Calculate the current in each resistor, but the 4 and 5ohm resistors arein parallel.
Given: E = 6V, r = 0.5, R1 = 3, R2 = 4, R3 = 5 ohms.
R = r + R1 + R2*R3/(R2+R3) = 0.5 + 3 + 2.22 = 5.72 ohms.
I = E/R = 6/5.72 = 1.05A = I1.
V1 = I1*R1 = 1.05 * 3 = 3.15 volts.
V2 = V3 = E - V1 = 6-3.15 = 2.85 volts.
I2 = V2/R2 = 2.85/4 =
I3 = V3/R3 =
Let R1=3,R=4,R=5
The first step is to calculate the effective resistance
For parallel connection
1/R=1/R2+1/R3
1/R= 1/4+1/5
R=2.22ohms
Series effective resistance
R= 2.22+3
R=5.22 ohms
Next, find the current flowing in the circuit
I= E/R+r
I=6/5.22+0.5
I=1.04A
Since the R1 is connected in series with the cell, same currents flow through it.
Current in R1=1.04A
P.d across R1= 1.04*3
P.d across R1=3.12V
Next, the lost voltage in the cell due to the internal resistance
Lost volt= I*r
Lost volt = 1.04*0.5
Lost volt=0.52ohms
V2=V3(connected in parallel)
V2=V3=E-V1-lost volt
V2=V3=6-3.12-0.52
V2=V3=2.36V
Current in R2=V2/R2
I2=2.36/4
I2=0.59A
I3= V3/R3
I3=2.56/5
I3=0.47A
I don't even understand anything
I did not get it
Tried a little
I apologize for the confusion. Is there anything you would like me to help you with?
Bulabalu
I'm sorry, I don't understand what you're asking for. Can you please provide more information or clarify your request?
total R = 0.5 Ω + 3 Ω + [4 * 5 / (4 + 5)] Ω
use total resistance and battery voltage to find the total circuit current
... this is the current in the 3 Ω resistor
... it is also the sum of the currents through the 4 Ω and 5 Ω resistors
current in a parallel array branch =
... (total circuit current) * [(total array resistance) / (branch resistance)]