Consider the circuit in the diagram below. in which r = 11 ω.

Consider the circuit in the diagram below, in which R = 11 ohm. (a) What is the resistance between points A and B? ______ Ohm(b) A 265-V emf is connected to the terminals A and B as shown above. What is the current in the R? ______________ A

Answer

The resistance R and 24 ohms are in paralle so using the
parallel equivalent of resistance we get (11*24)/(11+24) ohms
accross end of 15 ohms and the grnd of battry , i.e 7.543 ohms .
Now 7.543 ohms and 15 ohms are in series so using series equivalent
we have 15+7.543 ohms = 22.543 ohms accross terminals A and B.

the equivalent ckt looks like this

Now we find the current in our equivalent ckt which is simply
V/R using ohms law

V=265 V and R=22.543 ohms

=>I=V/R = 265/22.543 = 11.75 amperes.

Now we need to use KVL law, here we can say that the current in
15 ohms is saame as the current in equivalent resistance. The
current but divides at 11 ohms and 24 ohms .

since the resitors are parallel we have same voltage accross
both resistors.

let the current in 11 ohms be I1 and that in 24 ohms be I2

so we have I1 * 11 = I2 * 24 => I2= .458*
I1-----------------1

also we have I1+ I2=11.75-----------------2

use eqn 1 in 2

=> I1 + .458*I1 = 11.75

=> 1.458*I1 =11.75

=> I1 = 8.06 amperes

so the current in R is 8.06 amperes