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A battery of e.m.f 15V and internal resistance 3ohm is connected to two resistors 3ohm and 6ohm connected in parallel . Find:

  1. the  current through the battery,
  2. p.d between the terminals of the battery,
  3. the current in  3 ohm resistor,
  4. the current in  6 ohm resistor.

From the diagram,

External resistance 3 Ω and 6 Ω are in parallel. Equivalent resistance of these two is,

1/Rp = 1/3 + 1/6

=> Rp = 2 Ω

The equivalent resistance of the circuit is,

Req = 2 Ω + 3 Ω = 5 Ω

So, current through the battery is,

I = V/Req = 15/5 = 3 A

Now,

Voltage drop across the internal resistance is = 3 × 3 = 9 V

Thus, the potential difference across the terminals of the battery is = 9 V.

Thus the voltage drop across the external resistances that are in parallel is = 15 – 9 = 6 V

Thus, current through the external resistance 3 Ω is = 6/3 = 2 A

The current through the external resistance 6 Ω is = 6/6 = 1 A

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View Full Answer

(i) Rp(equivalent rersistence) in parallel  =>  1/Rp=1/R1+1/R2  

   1/Rp=1/3+1/6

    1/Rp=3/6   =>   Rp=2ohm

    now E=I(R+r)  => 15=I(2+3)  where I is current

    15=I x5   =>   I= 3A (Ans)

(ii) p.d between terminals of battery=Terminal Voltage

since E=V+v    where  V=Terminal Voltage & v=voltage drop

therefore first we have to find v   and since v=Ir 

v=3 x 3=9ohm

now since E=V+v  therefore V=15-9=6ohm (Ans)

(iii)since the voltage(i.e 6ohm ) in a PARALLEL CIRCUIT  is same everywhere 

therefore  current in 3ohm resistor :-

V=IR   => 6=I x 3  =>  I=6/3 or 2A (Ans)

(iv) current in 6ohm resistor:-

V=IR   =>   6=I x6   =>  I=6/6 i.e 1A  (Ans)

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