Answer to Question #142639 in Electric Circuits for Mary Rowe

Question

Two cells each having an emf of 1.5v and an internal resistance of 2 ohms are connected in parallel. Find the current in each case when the cells are connected to a 1 ohm’s resister

Answer 1

Here Two cells connected in parallel and then $1\varOmega$ connected in series is analysed completely. A circuit diagram based on the question is as follows,

There are multiple ways to analyse the circuit . Here I am using Thevenins theorm.

To find the thevenins equivalent of parallel connected cells, $1\varOmega$ resistor is removed and Open circuit voltage across is taken as Thevenins voltage $(V_{th})$ and resistance across, shorting the voltage sources is taken as Thevenins resistance $(R_{th})$ . i.e

From above circuit we have Thevenins voltage,

$V_{th} = 1.5V$ (Here as there will not be any current flow through the loop . Hence the 1.5 V will show up at output port).

From the above circuit Thevenins resistance ,

$R_{th} = 2//2 = \frac{2\times2}{2+2} = 1\varOmega$

Hence we have final thevenins circuit with the load resistor $1\varOmega$ as,

Hence the current through $1 \varOmega = \frac{1.5}{1+1} = 0.75A$

There fore the actual network is as follows,

As the load current is $0.75A$ Current through each cell $= 0.75/2 =0.375A$