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

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

• Diagram (Attachment provided in the question)

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}$

• Current flowing through the resistor.

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}$

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,

$\boxed{\pink{\sf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{1}{R_1} \: + \: \dfrac{1}{R_2}}}$

Here,

• $\sf R_1$ = 1.5 Ω

• $\sf R_2$ = 3 Ω

Substituting the values,

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

Taking LCM,

• $\sf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{2}{3} \: + \: \dfrac{1}{3}$

• $\sf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{3}{3}$

• $\sf \dfrac{1}{R_{Parallel}} \: = \: 1 Ω$

• $\sf R_{Parallel}$ = 1 Ω

Second,

• Finding the resistance in the 2nd set.

We know that,

$\boxed{\pink{\sf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{1}{R_1} \: + \: \dfrac{1}{R_2}}}$

Here,

• $\sf R_1$ = 6 Ω

• $\sf R_2$ = 12 Ω

Substituting the values,

• $\sf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{1}{6} \: + \: \dfrac{1}{12}$

Taking LCM,

• $\sf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{2}{12} \: + \: \dfrac{1}{12}$

• $\sf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{3}{12}$

• $\sf \dfrac{1}{R_{Parallel}} \: = \: \dfrac{1}{4}$

• $\sf R_{Parallel}$ = 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,

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

• Here,

• V = 9 V

• R = 5 Ω

Substituting the values,

• $\sf I \: = \: \dfrac{9}{5}$

• I = 1.8 A

Therefore,

• Current flowing through the resistors = 1.8 A.

$\Large{\bf{\purple{\mathfrak{\dag{\underline{\underline{Note:-}}}}}}}$

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

Answer 2

[Note - Omega from Web denotes Ohm.]

Given :-

Voltage = 9 V

To Find :-

Current  drawn from the battery

b. current flowing in the resistor

i. 1.5ohm

ii. 12ohm

Solution :-

We know that

Resistance for series = R1 + R2 + R3

Resistance for parallel = 1/R1 + 1/R2

ATQ

$\sf \dfrac{1}{R_n} = \dfrac{1}{1.5} + \dfrac{1}{3}$

$\sf\dfrac{1}{R_n} = \dfrac{1(2) + 1}{3}$

$\sf\dfrac{1}{R_n} = \dfrac{3}{3}$

$\sf R_n = 1 \Omega$

Now,

For the another resistor

$\sf \dfrac{1}{R_n} = \dfrac{1}{6} + \dfrac{1}{12}$

$\sf\dfrac{1}{R_n} = \dfrac{2(1) + 1}{12}$

$\sf\dfrac{1}{R_n} = \dfrac{2 + 1}{12}$

$\sf\dfrac{1}{R_n} = \dfrac{3}{12}$

$\sf\dfrac{1}{R_n} = \dfrac{1}{4}$

$\sf R_n = 4 \Omega$

Now

Therefore The total resistance = 4 + 1 = 5 $\Omega$

$\sf V = IR$

$\sf 9 = I(5)$

$\sf 9 =5I$

$\sf 1.8 = I$