__Question 38__
Two cells with electromotive forces

*E*1 and*E*2 and internal resistances*r*1 and*r*2 are connected to a resistor R as shown.
The terminal potential difference across cell 1 is zero.

Which
expression gives the resistance of resistor R?

**Reference:**

*Past Exam Paper – November 2019 Paper 12 Q34*

**Solution:**

**Answer: A.**

Let the current in the circuit
be I.

From Kirchhoff’s law,

Sum of e.m.f. in the
circuit = Sum of p.d. in the circuit

E

_{1}+ E_{2}= I (r_{1}+ r_{2}+ R)
R + r

_{1}+ r_{2}= (E_{1}+ E_{2}) / I eqn (1)
However, none of the
options available contains the current I. So, we need to substitute I by other
quantities.

Consider the cell 1.

e.m.f. of cell 1 =
terminal p.d. + lost volts

It is given that the
terminal p.d. is zero. So,

e.m.f. of cell 1 = lost
volts

E

_{1}= I × r_{1}
Current I = E

_{1}/ r_{1}
Replace the current I in
equation (1),

R + r

_{1}+ r_{2}= (E_{1}+ E_{2}) × r_{1}/ E_{1}
R + r

_{1}+ r_{2}= E_{1}r_{1}/E_{1}+ E_{2}r_{1}/E_{1}
R + r

_{1}+ r_{2}= r_{1}+ E_{2}r_{1}/E_{1}
R = r

_{1}+ E_{2}r_{1}/E_{1}– r_{1}– r_{2}
R = E

_{2}r_{1}/E_{1}– r_{2}
Note that r

_{2}can be written as E_{1}r_{2}/E_{1}so as to obtain the same denominator.
R = E

_{2}r_{1}/E_{1}– E_{1}r_{2}/E_{1}
R = (E

_{2}r_{1}–1r_{2}) / E_{1}