The figure below shows a part of a circuit. If a current of $12\,mA$ exists in the $5k\Omega$ resistor, find the currents in the other three resistors. What is the potential difference between the points $A$ and $B$?

Hint: In electrical circuits, when the resistors are connected in series combination, same current flows across each resistances while when they are connected in parallel combination the potential difference is same across each resistances, we will use this concept in order to find currents in each resistors and potential difference across point $A$ and $B$.

Formula used:
Net resistance between two resistances in series combination is calculated as ${R_{series}} = {R_1} + {R_2}$
Net resistance between two resistances in parallel combination is calculated as $\dfrac{1}{{{R_{parallel}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$
Current and potential difference across the resistor is related as $I = \dfrac{V}{R}$.

Complete step by step answer:
Firstly, let us find the net resistance among the parallel combination of resistances $20k\Omega (and)10k\Omega $ using the formula $\dfrac{1}{{{R_{parallel}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$ let net resistance among $20k\Omega (and)10k\Omega $ denoted as ${R_{20,10}}$ so we have,
$\dfrac{1}{{{R_{20,10}}}} = \dfrac{1}{{20}} + \dfrac{1}{{10}}$
$\Rightarrow \dfrac{1}{{{R_{20,10}}}} = \dfrac{{30}}{{200}}$
$\Rightarrow {R_{20,10}} = \dfrac{{20}}{3}k\Omega $

Now, we have three resistances which are connected in series with each other of resistances ${R_{20,10}} = \dfrac{{20}}{3}k\Omega $ , $5k\Omega $ and $100k\Omega $
now, it’s given that current of $12mA$ flows across the $5k\Omega $ resistance , and since three resistances of ${R_{20,10}} = \dfrac{{20}}{3}k\Omega $ , $5k\Omega $ and $100k\Omega $ are connected in series to each other which means same current of $12mA$ will flow across each resistance such that current across $100k\Omega $ will also be $12mA$. And current across ${R_{20,10}} = \dfrac{{20}}{3}k\Omega $ will be $12mA$,
Now, we have the resistance ${R_{20,10}} = \dfrac{{20}}{3}k\Omega $ is a parallel combination of $20k\Omega (and)10k\Omega $ so, net potential difference across ${R_{20,10}} = \dfrac{{20}}{3}k\Omega $ will be same in the individual resistances $20k\Omega (and)10k\Omega $.

Let ${V_{20,10}}$ is potential difference across the ${R_{20,10}} = \dfrac{{20}}{3}k\Omega $ with current of $12mA$ so, using $I = \dfrac{V}{R}$ we have,
${V_{20,10}} = \dfrac{{20}}{3}k\Omega \times 12mA$
$\Rightarrow {V_{20,10}} = 80V$
Now, current across the resistance $20k\Omega $ with potential difference of ${V_{20,10}} = 80V$ can be found as:
$I = \dfrac{{80}}{{20}}$
$\Rightarrow I = 4\,mA$
Similarly, the current in the $10k\Omega $ resistance with potential difference of ${V_{20,10}} = 80V$ can be found as:
$I = \dfrac{{80}}{{10}}$
$\Rightarrow I = 8\,mA$
Now, in order to find net potential difference between point A and B, we have the net current is flowing of $12mA$ in the series combination of resistances ${R_{20,10}} = \dfrac{{20}}{3}k\Omega $ , $5k\Omega $ and$100k\Omega $.

Let ${R_{AB}}$ denote the net resistance between point A and B then it’s calculated as:
${R_{AB}} = \dfrac{{20}}{3} + 100k\Omega + 5k\Omega $
$\Rightarrow {R_{AB}} = 121.67k\Omega $
Now, potential difference can be calculated by
${V_{AB}} = {R_{AB}}(12mA)$
$\Rightarrow {V_{AB}} = 121.67 \times 12$
$\therefore {V_{AB}} = 1460.04V$
Hence, the current in the other three resistances of $20k\Omega ,10k\Omega (and)100k\Omega $ is $4mA,8mA(and)12mA$ respectively.

The potential difference across the points A and B is ${V_{AB}} = 1460.04\,V$.

Note: It should be remembered that, the basic unit of conversions as $1mA = {10^{ - 3}}A$ which is a unit of current and the unit $1k\Omega = {10^3}\Omega $ which is a unit of resistance, and the formula between potential difference, current and resistance $I = \dfrac{V}{R}$ is known as the Ohm’s law. Resistance is the property of resistors to oppose the flow of current across them in the electrical circuits.