a battery with E=9V supplies current to the circuit. When the double-throw switch S is open as shown, the current in the battery is 1,5mA. When the switch is closed in position a,the current in the battery is 1,8mA. When the switch is closed in position b, the current is 2mA. Internal resistance of the battery is negligible.
1. Find the value of the resistance R1, R2 and R3 in the circuit
When the switch is open then the current goes through `R_1,R_2 ` and `R_3` are they are connected in series.
We are using the Ohm's law here.
`V = IR`
`9 = 1.5xx10^(-3)(R_1+R_2+R_3)`
`9/0.0015 = R_1+R_2+R_3------(1)`
When the switch is at 'a' the current goes through both `R_2` resistors top `R_3` and `R_1` . The `R_2` resistors are connected parallel. The resultant(`R'` ) of both` R_2` can be given as:
`1/R' = 1/R_2+1/R_2`
`R' = (R_2)/2`
Then `R'` and `R_1` are in series.
`9 = 1.8xx10^(-3)(R_1+R_2/2+R_3)`
`9/0.0018 = R_1+R_2/2+R_3 ------(2)`
When the switch is at 'b' the current goes through both `R_3` resistors top `R_2` and `R_1` . The `R_3` resistors are connected parallel. The resultant(R'') of both `R_2` can be given as:
`1/R'' = 1/R_3+1/R_3`
`R'' = (R_3)/2`
Then `R''` ,`R_2` and `R_1` are in series.
`9 = 2xx10^(-3)(R_1+R_2+R_3/2)`
`9/0.002 = (R_1+R_2+R_3/2) ----(3)`
`9/0.0018-9/0.0015 = -R_2/2`
`R_2 = 2000`
`9/0.002-9/0.0015 = R_2/2`
`R_3 = 3000`
`R_1 = 9/0.0015-2000-3000`
`R_1 = 1000`
So the answers are;
`R_1 = 1000ohm`
`R_2 = 2000ohm`
`R_3 = 3000ohm`