In the circuit shown below, $r_1$ and $r_2$ are resistors such that $\frac{r_1}{r_2}=4$ . If we connect a lamp as shown, the current through resistor $r_1$ increases by $\SI{0.1}{\ampere}$ .

Find the current through the lamp (in amperes).

The answer is 0.5.

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Relevant wiki: Potential Difference - A Classical Circuit - Kirchhoff´s LawBy applying Kirchoff's loop rule in the first loop, we get $i_{1}r_{1}+i_{2}r_{2}=V$ .

As $r_{1}, r_{2}$ and $V$ are constants, therefore, we can write $\Delta i_{1}r_{1}+ \Delta i_{2}r_{2}=0$ .

$\Delta i_2 = - \frac{\Delta i_1r_1}{r_2}.$ Since, $\dfrac {r_1}{r_2} =4$ , we get, $\Delta i_2 = - 4 \Delta i_1.$ Therefore, if the current in $r_{1}$ i.e. $i_1$ is increased by 0.1 $A$ , then $i_2$ will be decreased by 0.4 $A$ .

Now, if the current in $r_1$ and $r_2$ was $i$ before connecting the bulb, then after the connection.

$\begin{aligned} i_1 = i + 0.1 \\ i_2= i - 0.4. \end{aligned}$ The current through the bulb will be $i_{3} = i_{1}-i_{2} = \boxed{0.5 A}$ .