Circuit Solution via Incremental Quantities

Electricity and Magnetism Level 3

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.

2 solutions

Rohith M.Athreya
Jan 17, 2017

Relevant wiki: Potential Difference - A Classical Circuit - Kirchhoff´s Law

By 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 $i_{3} = i_{1}-i_{2} = \boxed{0.5 A}$ .

I and Agnishom discussed your solution in the slack. We both agree that the solution is great but not presented in an elaborate manner and is difficult to understand.
As per our discussions, the solution can be improved as follows,
-I would tell about how I wrote the equations like i'd say by applying Kirchoff's loop rule in the first loop we get $i_1r_1 + i_2r_2 = V$ .
- I'd also elaborate on the intermediate steps like "As the $r_1$ , $r_2$ and $V$ are constant, therefore, we can write $\Delta i_1r_1 +\Delta i_2r_2 = 0$
- After finding the relation between $\Delta i_1$ and $\Delta i_2$ I would further explain that if the current is increased in $r_1$ is increased by 0.1 A then it will decrease by 0.4A in the $r_2$ /. Hence, the current in the bulb will be $i_3 = i_1 - i_2$ .
- you may use the following diagram

Rohit Gupta - 4 years, 4 months ago

done

thanks

Rohith M.Athreya - 4 years, 4 months ago

Rohit Sachdeva
Jan 19, 2017

Let initially the current in the loop be $I$ .

Ofcourse $V = 5Ir$ ......[1]

Let resistance of lamp be $R$ . Let current in the loop be $i$ which gets split into $i_1$ through lamp and $i-i_1$ through resistor $r$ . Since it is given that

$i = I+0.1$ , from [1] we get,

$V = 5(i-0.1)r = 5ir-0.5r$ ......[2]

Applying Kirchhoff's law in left loop we get,

$V = 4ir+(i-i_1)r = 5ir-i_1r$ ......[3]

Comparing [2] and [3], $i_1 = 0.5 A$

Circuit Solution via Incremental Quantities

The answer is 0.5.

2 solutions

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