Question

7. In the given circuit, calculate the:
current drawn from the battery
b. current flowing in the resistor
i. 1.5ohm
ii. 12ohm​

Answer 1

$\sf\small{ \underline{\underline\red{ \pmb{Given:-}}}}$

• Diagram (Attachment provided in the question)

$\sf\small{ \underline{\underline\red{ \pmb{ To \: prove:- }}}}$

• Current flowing through the resistor

$\sf\small{ \underline{\underline\red{ \pmb{ Solution:- }}}}$

Here,

We can observe that,

• There are 2 set of resistors connected in series and in that each set the resistors are connected in parallel.

First,

Finding the resistance in the 1st set,

We know that,

$\boxed{\purple{\bf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{1}{R_1} \: + \: \dfrac{1}{R_2}}}$

Here,

$\sf R_1= 1.5 Ω \\ \\ \sf R_2 = 3 Ω$

Substituting the values,

$\bf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{1}{1.5} \: + \: \dfrac{1}{3}$

Taking LCM,

$\bf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{2}{3} \: + \: \dfrac{1}{3} \\ \\ \\ \bf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{3}{3} \\ \\ \\ \bf \dfrac{1}{R_{Parallel}} \: = \: 1 Ω \\ \\ \\ \bf R_{Parallel}$

Second,

Finding the resistance in the 2nd set.

We know that,

$\boxed{\purple{\bf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{1}{R_1} \: + \: \dfrac{1}{R_2}}}$

Here,

$\sf R_1 = 6 Ω \\ \\ \sf R_2 = 12 Ω$

Substituting the values,

$\bf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{1}{6} \: + \: \dfrac{1}{12} \\ \\ \bf \: Taking \: LCM, \\ \bf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{2}{12} \: + \: \dfrac{1}{12} \\ \\ \bf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{3}{12} \\ \\ \bf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{1}{4} \\ \\ \bf R_{Parallel}= 4 Ω$

Now,

Since the resistors are connected in series,

Total resistance = Resistance of set 1 + Resistance of set 2.

Substituting the values,

Total resistance = 1 Ω + 4 Ω

Total resistance = 5 Ω

Now,

(a) Current drawn from the battery =

(b) Current flowing through the resistor =

We know that,

$\boxed{\purple{\sf I \: = \: \dfrac{V}{R}}}$

Here,

V = 9 V

R = 5 Ω

Substituting the values,

$\sf I \: = \: \dfrac{9}{5}$

Therefore,

Current flowing through the resistors = 1.8 A.

$\sf\small{ \underline{\underline\red{ \pmb{Note:-}}}}$

• Current drawn from the battery and current flowing through the resistors both mean the same.