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Question

# A circuit with resistors, a cell and an ammeter is shown in the figure. Find the value of current coming out from the battery and also find the ammeter reading.

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Solution

## The circuit in the question can be simplified to figure 1.1, which can be further be simplified as figure 1.2. From the figure 1.2, we can see, that the three 4Ω resistors are in parallel and this parallel combination is in series with the other 4Ω resistor and the battery. The Equivalent resistance of the parallel combination is given by1Req=14+14+14=34⇒Req=43Ω This simplifies the circuit as shown in figure 1.3Now, the Req and the 4Ω resistor are in series. So, the equivalent resistance for this combination is given by,Rs=Req+4=43+4=163ΩSo the circuit simplifies to figure 1.4Using Ohm's Law, we haveV=iR2=i×163⇒2163=i⇒i=38ANow, the current in the ammeter will be i3A because the current entering the parallel branch is i and since all the branches are having equal resistances, the current divides equally in all the branches. Thus, the ammeter reading will be iA=13×38=18AHence, the current in the circuit, i is 38A and the ammeter reading is 18A.

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