Question

10 ohms and 20 ohms resistors are combined in series. Calculate the effective resistance. Also calculate the current flowing when the combination is connected to 6 volts battery. Also find the potential across 10 ohms and 20 ohms resistors separately

Answer 1

Answer:

(1) Current flowing when connected together

$R_{1}$=10Ω

$R_{2}$=20Ω

V=6 V

$R_{S}$=$R_{1}$+$R_{2}$=10+20=30Ω

I=$\frac{V}{R}$=$\frac{6}{30}$=0.2Ω

(2)Current flowing through resistors separately

Since the resistors are in series, the current flowing through them is equal

a. $R_{1}$=10Ω

I=0.2Ω

V=IR=0.2×10=2 V

b.$R_{2}$=20Ω

V=IR=0.2×20=4 V

Answer 2

10 ohms and 20 ohms resistors are combined in series.

Given that, R1 = 10 Ω and R2 = 20 Ω

Now,

In series, Resistance = Sum of all the resistors connected in series.

In question, we have given two resistors connected in series.

Rs = R1 + R2

Rs = (10 + 20) Ω = 30 Ω

At first, we have to calculate the effective resistance, which is done above and Effective resistance is 30 Ω.

Now, we have to calculate the current flowing when the combination is connected to 6 volts battery.

Given that, V = 6V and R = 30 Ω

We know that,

V = IR

I = V/R

I = 6/30 = 1/5 = 0.2 A

And at last, we have to find the potential across 10 ohms and 20 ohms resistors separately.

For 10 ohms

V = IR = 0.2 × 10 = 2 V

For 20 ohms

V = IR = 0.2 × 20 = 4 V