Question

Please solve Q.12...I will mark brainliest.​

Answer 1

Answer:total resistance= 9.2ohm

Total current= 0.48 ampere

Step-by-step explanation:

Please mark it as the brainliest answer if found helpful

Answer 2

Given that,

Potential Difference (V) = 4.5V

Resistance of five resistors = $3.6 \Omega, 6 \Omega, 3 \Omega, 1 \Omega, 3 \Omega$

To find,

Find total resistance and total current of the circuit = ?

Assumption,

Let the five resistors whose resistances are $3.6 \Omega, 6 \Omega, 3 \Omega, 1 \Omega, 3 \Omega$ be R1, R2, R3, R4 ans R5 respectively.

As we can see that,

Three resistors are connected in series and two resistors are in parallel.

As we know,

Equivalent resistance of resistors connected in parallel is,

$\fbox{\bold{R_s = R_1 + R_2 + R_3 .... R_n}}$

Here, the equivalent resistance for three resistors which are connected in series is,

Rs = R1 + R2 + R3

Rs = $3.6 \Omega + 6 \Omega + 3 \Omega$

$\fbox{\bold{R_s = 12.6 \Omega}}$

So, the equivalent resistance of resistors which are connected in series is $12.6 \Omega$

Now, two resistance are connected in parallel, so it's equivalent resistance for n number of resistors is,

$\mathsf{\fbox{\bold{\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} ..... \frac{1}{R_n}}}}$

But the equivalent resistance for 2 resistors which are connected in parallel is,

$\mathsf{\frac{1}{R_p} = \frac{1}{R_4} + \frac{1}{R_5}}$

$\mathsf{\frac{1}{R_p} = \frac{1}{1 \Omega} + \frac{1}{3 \Omega}}$

$\mathsf{\frac{1}{R_p} = \frac{3 + 1}{3}}$

$\mathsf{\frac{1}{R_p} = \frac{4}{3}}$

$\mathsf{R_p = \frac{3}{4}}$

$\mathsf{\boxed{\bold{R_p = 0.75 \Omega}}}$

So, the equivalent resistance of 2 resistors which are connected in parallel is, $\mathsf{\bold{R_p = 0.75 \Omega}}$.

So, the total resistance of the circuit is,

Total resistance = equivalent resistance in series + equivalent resistance in parallel

Total resistance = $12.6 \Omega + 0.75 \Omega$

$\mathsf{\fbox{\bold{Total\:resistance = 13.35 \Omega}}}$

For the current:

As we know, the current flowing through a series of resistors is constant and the current flowing through resistors which are connected in parallel is inconstant.

So, the current flowing through resistors of which are connected in series is,

Current (I) = $\mathsf{\frac{V}{R_s}}$

Current (I) = $\mathsf{\frac{4.5}{12.6}}$

$\fbox{\bold{Current (I) = 0.36 A}}$ (approx)

So, the current flowing through the circuit with resistors which are connected in is approximately 0.36 A.

Now, the current flowing through the parallel resistors is different at every resistance.

So, the current flowing through R4 is,

Current (I) = $\mathsf{\frac{V}{R_4}}$

Current (I) = $\mathsf{\frac{4.5}{1}}$

$\mathsf{\fbox{\bold{Current\:(I) = 4.5\:A}}}$

So, the current flowing through R5 is,

Current (I) = $\mathsf{\frac{V}{R_5}}$

Current (I) = $\mathsf{\frac{4.5}{3}}$

$\mathsf{\fbox{\bold{Current\:(I) = 1.5 A}}}$

So, the current flowing through R5 is 1.5 A.

Now, the total current flowing through the parallel resistors = current in R4 + current in R5

Total current = 4.5 + 1.5

$\fbox{\mathsf{\bold{Total\:current = 6\:A}}}$

Now, the equivalent current flowing throughout the circuit = total current in series + total current in parallel

Total current = 0.36 + 6

$\mathsf{\fbox{\bold{Total\:current = 6.36\:A}}}$

So, the current flowing through the circuit is 6.36 A.