Question

A parallel combination of an 4 resistor and an unknown resistor R is connected in series with a 19 resistor and a battery. This circuit is then disassembled and the three resistors are then connected in series with each other and the same battery. In both arrangements, the current through the 4 resistor is the same. What is the unknown resistance R?

Answer #1

So in the series circuit the voltage of the battery is R is the
unknown resistor):

I ( R + 23)

And in the first circuit the current through R is 4I/R (since the
voltage across this resistor is 4I (same as in the 4 ohm
resistor).

Then the voltage of the battery in this first circuit is given
by:

4I + 19(I + 4I/R)

If you equal both equations given the battery voltage you can
eliminate I and solving -

IR+23I= 4I+19I+76I/R

R+23=4+19+76/R

R.R=76

you will find that the unknown resistor is sqrt(76) ohms= 8.71 ohm

Ans- R= 8.71 ohm

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