Question

Two resistances of 3 ohms and 2 ohms are connected across a 1.5 volt battery. Calculate the current in the main circuit, when the resistances are in
(i) series,
(ii) parallel.​

Answer 1

$\huge {\underline{\pink{Given:-}}}$

• Resistance of Resistor ,R1 = 3Ω

• Resistance of Resistor ,R2 = 2Ω

• Potential Difference ,V = 1.5v

$\huge {\underline{\green{To Find:-}}}$

• Current in Series
• Current in Parallel

$\huge {\underline{\blue{Solution:-}}}$

Firstly we calculate the Equivalent Resistance in Both Case

(i) Series

Equivalent Resistance in series is the sum of resistance of each Resistors.

• Req = R1 + R2 + ....

Substitute the value we get

➱ Req = 3Ω + 2Ω

➱ Req = 5Ω

Therefore, the Equivalent resistance in series is 5 ohms.

Now, we have to calculate the Current So ,by using Ohm's Law

• V = IR

Where

V is the Potential Difference

I is the current

R is the Resistance

Substitute the value we get

➱ 1.5 = I × 5

➱ I = 1.5/5

➱ I = 0.3 A

Therefore, the current through the resistor when they are connected in series is 0.3 Ampere.

____________________________________

(ii) Parallel

The Equivalent resistance in the parallel circuit is the sum of reciprocal of resistance of each esistors.

• 1/Req = 1/R1 + 1/R2 + ...

Substitute the value we get

➱ 1/Req = 1/3 + 1/2

➱ 1/Req = 2+3/6

➱ 1/Req = 5/6

➱ Req = 6/5

➱ Req = 1.2 Ω

Therefore ,the equivalent resistance when Resistors are connected in Parallel is 1.2

Using Ohm's Law

• V = IR

Substitute the value we get

➱ 1.5 = I × 1.2

➱ I = 1.5/1.2

➱ I = 15/12

➱ I = 1.25 A

Therefore, the Current through the circuit when the Resistor are connected in parallel is 1.25 Ampere .

Answer 2

Given:-

• Resistance, R₁ = 3Ω
• Resistance, R₂ = 2Ω
• Potential Difference, V = 1.5V

Find:-

• Current flowing when the resistance are in series.
• Current flowing when the resistance are in parallel.

Solution:-

(i)

When, connection of resistance are in series.

$\ast\sf R = R_1 + R_2$

$\sf where \small{\begin{cases} \sf R_1 = 3\Omega \\ \sf R_2= 2\Omega \end{cases}}$

$\implies\sf R = 3 + 2$

$\implies\sf R = 5\Omega$

Now, using Ohm's Law

$\ast\sf V = I R$

$\ast\sf I = \dfrac{V}{R}$

$\sf where \small{\begin{cases} \sf V = 1.5V\\</p><p>\sf R = 5\Omega</p><p> \end{cases}}$

$\implies\sf I _{series} = \dfrac{1.5}{5}$

$\implies\sf I _{series} = 0.3 A$

$\boxed{\small{\therefore \underline{\textsf{Current flowing in series is 0.3A}}}}$

$\qquad$________________________

(ii)

When, connection of resistance are in parallel.

$\ast\sf \dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$

$\sf where \small{\begin{cases} \sf R_1 = 3\Omega \\ \sf R_2= 2\Omega \end{cases}}$

$\implies\sf \dfrac{1}{R} = \dfrac{1}{3} + \dfrac{1}{2}$

$\implies\sf \dfrac{1}{R} = \dfrac{2 + 3}{6}$

$\implies\sf \dfrac{1}{R} = \dfrac{5}{6} \Omega$

$\implies\sf R= \dfrac{6}{5} \Omega$

Now, using Ohm's Law

$\ast\sf V = I R$

$\ast\sf I = \dfrac{V}{R}$

$\sf where \small{\begin{cases} \sf V = 1.5V\\</p><p>\sf R = \dfrac{6}{5}\Omega</p><p> \end{cases}}$

$\implies\sf I_{parallel} = \dfrac{1.5}{ \dfrac{6}{5}}$

$\implies\sf I_{parallel} = \dfrac{15}{10} \times \dfrac{5}{6}$

$\implies\sf I_{parallel} = \dfrac{75}{60}$

$\implies\sf I_{parallel} = \dfrac{15}{12}$

$\implies\sf I_{parallel} = \dfrac{5}{4}$

$\implies\sf I_{parallel} = 1.25A$

$\boxed{\small{\therefore \underline{\textsf{Current flowing in Parallel Combination is 1.25A}}}}$