Question
Question: The current drawn \(5V\) source in the given circuit will be:  resistor and we can remove it. Now, the remaining resistances can be solved by either in series combination or parallel combination of resistors.
Series equivalent of 10Ω and 20Ω : 10+20=30Ω
Series equivalent of 5Ωand 10Ω : 5+10=15Ω
Now, these two resistances, 30Ωand 15Ωare in parallel. So, their parallel equivalent :
Req=30+1530×15=45450=10Ω
Hence, the equivalent resistance of the circuit is 10Ω.
Now, let us use Ohm’s law to find the current in the circuit.
V=IR⇒I=RV
⇒I=105=0.5A
Hence, option (B) 0.5A is correct.
Note:
Whenever a combination of 5 resistance is present and they are neither in parallel combination nor in series combination. We will use the Wheatstone bridge concept.
The equivalent of the series combination is the algebraic sum of all resistances.
Req=R1+R2
The equivalent of the parallel combination is given by the following formula.
Req=R1+R2R1R2