In: Physics
We will analyse the current in each circuit one by one,
Circuit 1
\(V=\varepsilon-I R_{e f f}\)
\(9=10-I * 10\)
\(I=0.1 A\)
Circuit 2
\(V=\varepsilon-I R_{e f f}\)
\(9=10-I * 20\)
\(I=0.05 A\)
Circuit 3
\(V=\varepsilon-I R_{e f f}\)
The 2 resistors are in parallel,
10 ohm parallel with 10 ohm
effective resistance when two resistors are in parallel is given by,
\(R_{e f f}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}\)
\(9=10-I * \frac{10 * 10}{10+10}\)
\(I=0.2 A\)
Circuit 4
\(V=\varepsilon-I R_{e f f}\)
The 3 resistors are in parallel,
10 ohm parallel with 10 ohm gives equivalent resistance of \(5 \mathrm{ohm}\), this 5 ohm is parallel with \(50 \mathrm{ohm}\)
effective resistance when two resistors are in parallel is given by,
\(R_{e f f}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}\)
\(9=10-I * \frac{5 * 50}{50+5}\)
\(I=\frac{1}{4.55}=0.22 A\)
Therefore the descending order of total current coming from the battery,
Circuit \(4>\) Circuit \(3>\) Circuit \(1>\) Circuit 2