Question

1. Find the equivalent resistance and the amount of current drawn from
battery:
​

Answer 1

A all 4 resistor are in parallel >b<

Answer 2

Solution :

Please refer the attachment first

Start solving the circuit from right to left

R 1 and R2  having resistance of  15 Ω and 10 Ω are connected in parallel .

Equivalent resistances for n no of resistors is connected in parallel is given by ,

➜   1 / Rp = 1 / R 1 + 1 / R 2 .... + 1 / Rn

Hence ,

➜  1 / Rp = 1 / R 1 + 1 / R2

➜  1 / Rp = 1 / 15 + 1 / 10

➜  1 / Rp = 2 + 3 / 30

➜  1 / Rp = 5 / 30

➜  Rp = 6 Ω

Rp , R 3 and R 4 are connected in series with each other ,

Equivalent resistance of n no of resistors connected in series with each others .

➜  Rs = R 1 + R 2 ... + R n

Hence ,

➜  Rs = R p + R 3 +  R 4

➜  Rs = 6 + 2 + 7

➜  Rs = 15 Ω

Rs is in parallel with R 5 and R6 each having a resistance of 5  Ω  and both the 5  Ω  resistors are connected in series with each other

➜  Rs '= R 5 + R 6

➜  R s ' = 5 + 5

➜  R s ' = 10  Ω  

➜  1 / Rp ' = 1 / R s   + 1 / Rs'

➜  1 / Rp ' = 1 / 15 + 1 / 10

➜  1 / Rp ' = 2 + 3 / 30

➜  Rp ' = 30 / 5

➜  Rp ' = 6 Ω

Rp' and R 7 are connected in series with each other

➜  R s " = Rp ' + R7

➜  Rs " = 6 + 4

➜  R s " = 10 Ω  

Equivalent resistance of the circuit = 10 Ω  

➜  Voltage = 25 V

Current drawn (I)

By using Ohm's law ,

➜  I = V / Rs"

➜  I = 25 / 10

➜  I = 2.5 A

The current drawn from the circuit is 2.5 A