Question

Two cells each of emf 2v and internal resistance 0.5 ohms are connected in series. They are made to supply currents to a combination of three resistors, one of the resistance 2ohms is made is connected in series to a parallel combination of two other resisters each of resistance 3 ohms.

Draw the circuit diagram and calculate.

i. Current in the circuit

ii. The potential difference across the parallel combination of the resistor

iii. Lost voltage of the batting.

Answer 1

Answer:-

we can calculate equivalent resistance of the circuit = $2+\frac{3}{2}+ 0.5+0.5=\frac{9}{2}$ ohm

also equivalent emf when voltage source is in series = $V_1+V_2=2+2=4 \ Volt$

i) by using formula $I = \frac{V}{R}=\frac{4}{9\over 2}=\frac{8}{9}=0.88 \ amp$

ii) Potential difference across the parallel combination of the resistor

$V-IR=4-0.88\times (2+0.5+0.5)ohm=1.36\ volt$

iii) Lost voltage

$V=IR=0.88(0.5+0.5)=0.88 \ volt$