Resistances R_{1,} R_{2}, R_{3} and R_{4} are connected as shown in the figure. S1 and S2 are two keys. Discuss the current flowing in the circuit in the following cases.

a. Both S_{1} and S_{2} are closed.

b. Both S_{1} and S_{2} are open.

c. S_{1} is closed but S_{2} is open.

#### Solution

**a. **The circuit for this case is shown below.

Here, R_{1} and R_{2} are in parallel. Their effective resistance is

`R' = (R_1 xx R_2)/(R_1 + R_2)`

Also, R_{4} is in parallel with a conductor of 0 resistance. Their equivalent resistance is therefore 0. The given circuit thus becomes as shown below.

Now , R and `R_3` are in series . Thus , the quivalent resistance of the circuit is

`R_(eq) = (R_1 xx R_2)/(R_1 + R_2) + R_3 = (R_1R_2 + R_1R_3 + R_3R_2)/(R_1+R_2)`

Hence, the current flowing in the circuit is

`I = V/((R_1R_2 + R_1R_3 + R_3R_2)/(R_1 + R_2))` = `(V(R_1 + R_2))/(R_1R_2 + R_1R_3 + R_3R_2)`

where , V is the potential difference of the battery . **b.**The circuit for this case is shown below.

Here, R_{1}, R_{4} and R_{3} are in series. Therefore, the equivalent resistance of the circuit is

`R_(eq) = R_1 + R_2 + R_3`

Hence, the current flowing in the circuit is

`I = V/R_(eq) = V/(R_1 + R_2 + R_3)`

where, V is the potential difference of the battery.**C.**The circuit for this case is shown below.

Here , R_{1} and R_{2} are in parallel . their effective resistance is

`R' = (R_1 xx R_2)/(R_1 + R_2)`

Now , R_{3} , R and R_{4} are in series . Thus , the equivalent resistance of the circuit is

`R_(eq) = (R_1 xx R_2)/(R_1 +R_2) + R_3 + R_4` = `(R_1R_2 + R_1R_3 + R_3R_2 + R_1R_4 + R_4R_2)/(R_1 + R_2)`

Hence , the current flowing in the circuit is

`I = (V/(R_1R_2 + R_1R_3 + R_3R_2 + R_1R_4 + R_4R_2))/(R_1 + R_2) = (V(R_1 + R_2))/(R_1R_2 + R_1R_3 + R_3R_2 + R_1R_4 + R_4R_2)`

where , V is the potential difference of the battery.