Answer:
Explanation:
Given that,
Battery capacity
E=12V
Internal resistance of battery r=4ohms
We are told that the circuit must have at least 2 resistor
If the terminal voltage is Vt= 11.478V
Let all the resistor we want to connect the battery have a equivalent resistance Req
Req = R1 +R2 + R3+...+ Rn
Series connection
Using KVL in the circuit,
E = I(R+r)
12 = I ( Req + 4) equation 1
When the terminal voltage is 11.478V, I.e voltage across the external resistance, in this case the equivalent resistance Req
Using ohms law
Vt=I•Req
Then,
11.478 = I•Req equation 2
Make I subject of formulae
Then, I = 11.478/Req. Equation 3
Substitute equation 3 into 1
12 = I ( Req + 4)
12 = 11.478/Req ( Rea + 4)
12 = 11.478 + 45.912/Req
12 - 11.478 = 45.912/Req
0.522 = 45.912/Req
Cross mutilply
0.522 Req = 45.912
Then, Req = 45.912/0.522
Req = 87.95 ohms
Req ≈ 88 ohms
The resistor must have a resistance of 88 ohms and they have a series configuration
