91Ó°ÊÓ

1. Potential$V_A=12V$
2. Potential difference$\mathrm{Δ}{V}_B=2Ω$
3. Potential differencerole="math" $\mathrm{Δ}{V}_C=5Ω$
4. Resistance$r_3=200Ω$

Using the property of parallel circuit, find the voltagedrop across resistor 3. Inserting it in the Ohm’s law, find the current in the circuit.Also, apply Ohm’s law to resistor 1 and resistor 2 and can find the values of their resistances.

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

Formulae are as follow:

$I=\frac{V}{R}$

Where, I is current, V is voltage, R is resistance.

The resistance of resistor 1:

Since, the resistances in upper and lower branch are parallel, the voltage drop across both branches is equal.

So, the voltage drop across the upper branch is 12 V.

Hence, the voltage drop across the resistor 3 is 5V and the current in the circuit is given by,

$$\begin{gathered} i=\frac{\mathrm{Δ}V}{r_3} \\ i=\frac{5V}{\left(200\mathrm{Î\copyright }\right)} \\ i=25mA \end{gathered}$$

Then the resistance of resistor 1 will be,

$$\begin{gathered} i=\frac{\mathrm{Δ}V}{r_1} \\ {r}_1=\frac{2V}{25mA} \\ {r}_1=80Ω \end{gathered}$$

Hence, the resistance of Resistor 1 is$r_1=80\mathrm{Î\copyright }$

The resistance of resistor 2 :

From the graph, we can see that the resistor 2 has the same voltage drop as the resistor 3 and is given by,

$\mathrm{Δ}V=5V$

Its resistance is then given by,

$$\begin{gathered} i=\frac{\mathrm{Δ}V}{r_2} \\ {r}_2=\frac{\mathrm{Δ}V}{i} \\ {r}_2=\frac{5V}{25mA} \\ {r}_2=200Ω \end{gathered}$$

Hence, the resistance of Resistor 2 is$r_2=200\mathrm{Î\copyright }$

Therefore, by using ohms law the resistance of the resistors can be determined.