Question

For the circuit shown in the given diagram: What is the value of
(i) current through $6\Omega $ resistor?
(ii) potential difference across $12\Omega $ resistor?

Answer 1

Hint: This question is solved by the equation of ohm's law. So, students should know what ohms law is, what is a resistor and what is the potential difference. Ohm’s law gives the relation between current passing and voltage applied across any circuit.

Complete step by step answer:
Ohm’s Law states that the current flowing through a conductor is directly proportional to the potential difference applied across its ends, provided the temperature and other physical conditions remain unchanged.

As per ohm’s law Current is directly proportional to voltage difference through a resistor. That is, if the current doubles, then so does the voltage. To make a current flow through a resistance there must be a voltage across that resistance. Ohm's Law shows the relationship between the voltage (V), current (I) and resistance (R)
$I\propto V$
$ \Rightarrow V = IR$
Where R is a constant called resistance of the conductor. The value of this constant depends on the nature, length, area of cross section and temperature of the conductor.

A resistor is a passive two-terminal electrical component that implements electrical resistance as a circuit element. Resistors are used to reduce current flow, adjust signal levels, to divide voltages etc in an electrical circuit.Let the current flowing through the circuit be I and it is divided into 2 in which one flows through branch AB and the other through DC.

Let the current flowing through AB be ${I_1}$ and current flowing through DC be ${I_2}$
Therefore,
$I = {I_1} + {I_2}$
The total resistance in the arm AB
${R_{AB}} = {R_1} + {R_2}$
$\Rightarrow {R_{AB}}{\text{ = 6}}\Omega {\text{ + 3}}\Omega {\text{ = 9}}\Omega $
And in the arm CD
${R_{DC}} = {R_3} + {R_4}$
$\Rightarrow {R_{DC}} = 12\Omega + 3\Omega = 15\Omega $

(i) Now the current flowing through in the resistor $6\Omega $ is
${I_1} = \dfrac{V}{{{R_{AB}}}}$
By substituting the values, we get
${I_1} = \dfrac{{4V}}{{9\Omega }}$
$\therefore {I_1} = 0.44{\text{ A}}$

(ii) Now the current flowing through DC
${I_2} = \dfrac{V}{{{R_{DC}}}}$
By substituting the values, we get
${I_2} = \dfrac{{4{\text{ V}}}}{{15{\text{ }}\Omega }}$
$\Rightarrow {I_2}{\text{ = 0}}{\text{.27 A}}$
Therefore, now the potential difference across the resistor $12\Omega $ is
${V_3} = {I_{DC}} \times {R_3}$
$\Rightarrow {V_3} = 0.27{\text{ A }} \times {\text{ 12 }}\Omega $
Further simplifying, we get
$\therefore {V_3} = 3.24{\text{ V}}$

Note: The SI unit of resistance is Ohm which is denoted by the symbol Omega $\left( \Omega \right)$. Students should be careful while writing units after solving the problems. The answer should be written with proper unit’s Students shouldn’t forget to write the units along with the answer. All the physical quantities should be written along with the proper units. And the ohms law is not applicable for non ohmic devices and non-linear devices.