Question

three resistors of 13ohom, 17 ohom & 10 ohom are connected in series. combination. voltes 30 then find current ​ ​

Answer 1

Given :-

★ Three resistors of 13 Ω, 17Ω and 10 Ω are connected in series.

★ Potential difference of 30 volts is applied through the combination.

To Find :-

★ The current flowing through circuit.

Solution :-

We know,

the equivalent Resistance is given by

Rs = R1 +R2 +R3 +......... +Rn

According to the question :-

R1 = 13 Ω

R2 =17 Ω

R3 =10 Ω

Substituting the given values, we get

Rs = R1 +R2 +R3

⟹ Rs =( 13 +17 +10) Ω

⟹ Rs = 40 Ω

☞ Ohm's law

V =IR

Where,

V = Voltage

R = Resistances

I = Current

Now, put the given values

I = 30/40

$⟹I = 0.75 \: A$

Hence,

the value of current flowing through circuit is = 0.75 A

Know More :-

✰What is series combination?

=> When current flow only one direction not change, it said to be series combination.

✰ What is parallel combination?

=> When current flow different direction, it said to be parallel combination.

Answer 2

Given ,

The three resistor of 13 ohm , 17 ohm and 10 ohm are connected in series with 30 v

We know that , the equivalent resistance in series combination is given by

$\boxed{ \tt{R = R_{1} +R_{2} + R_{3} + ... }}$

Thus ,

R = 13 + 17 + 10

R = 40 ohm

Now , the relationship between current and potential difference is given by

$\boxed{ \tt{V = IR}}$

Thus ,

30 = I × 40

I = 30/40

I = 0.75 Amp

Hence , The current is 7.5 Amp