Question

16. Three resistors are connected as shown in the diagram.

Through the resistor 5 ohm, a current of 1 ampere is flowing.
(i) What is the current through the other two resistors ?
(ii) What is the p.d. across AB and across AC?
(iii) What is the total resistance ?
17
Con+b
in the diramhalom.​

Answer 1

Answer:

(i)0.6A, 0.4A respectively

(ii) 5V and 11 V respectively

(iii) 11Ω

This is a lengthy but easy one. Try to understand. Hope it helps.

Explanation:

Let us find the equivalent resistance of the circuit.

Let R₁ = 5Ω , R₂ = 10Ω  and R₃ = 15Ω

R₂ and R₃ are in parallel.

⇒1/R(eff) = 1/R₂ + 1/ R₃

              = 1/10 + 1/15

              = 1/5(1/2 + 1/3)

              = 1/5*(5/6)

              = 1/6

∴R(eff) = 6Ω

Now R(eff) and R₁ are in series.

∴R(eq) = R(eff) + R₁

          = 6 + 5

          = 11Ω

(iii) Total resistance = 11Ω

From Ohm's law

V = iR        [i = 1A]

⇒V = 1*11      [Here R is R(eq) of the circuit]

∴V = 11 V   ________(a)

Voltage drop or potential drop at 5Ω resistor.

From Ohm's law

V = i*R

⇒V = 1*5

∴ V = 5V  _______(b)

Out of 11 volts, 5 volts are consumed by a 5Ω resistor

So the remaining 6 volts are consumed by the parallel connection.

The other two resistors are in parallel.

So voltage remains constant but current varies

Let i = 1A splits into currents i' and i''.

⇒i' + i'' = 1A

V = 6 volts

1. current i' flows through the resistor of resistance 10Ω

  V = 6 volts (remain constant in parallel connection)

  From Ohm's law

  V = iR

⇒6 = i'*10

⇒i' = 6/10 = 0.6A

∴ i' = 0.6A _____(1)

2. current i'' flows through the resistor of resistance 15Ω

   V = 6 volts (remain constant in parallel connection))

   From Ohm's law

   V = iR

⇒6 = (i'')*15

⇒i'' = 6/15 = 0.4

∴i'' = 0.4A ______(2)

(i) So the current flowing through the resistor of resistance

1. 10Ω is i' = 0.6A
2. 15Ω is i'' = 0.4A

(ii) Potential difference across AB = $V_{B}-V_{A}$ = 5 V   [from (b)]

    Potential difference across AC = $V_{C}-V_{A}$ = 11 V  [from (a)]