In the figure shown, what is the current (in Ampere) drawn from the battery ? you are given
R1=15Ω,R2=10Ω,R3=20Ω,R4=5Ω,R5=25Ω,R6=30Ω,E=25V
13/24
7/18
9/32
15/32
Apply series and parallel combination of resistors 20Ω, 5Ω and 25Ω are in series ⇒R1=20+5+25=50Ω ⇒R1=50Ω is parallel with 10Ω ⇒R2=50(10)50+10=50060=253Ω ⇒R2=253Ω is in series with 15Ω and 30Ω Reff=253+15+ 30=1603Ω i=VoltageNet resistance of the circuit=251603=1532