1.

The galvanometer shown in the figure has resistance 10Omega. It is shunted by a series combination of a resistance S = 1Omega and an ideal cell of emf 2 V. A current 2 A passes as shown.

Answer»

The READING of the galvanometer is 1A.
The reading of the galvanometer is zero.
The potential difference across the resistance S is 1.5 V.
The potential difference across the resistance S is 2 V.

Solution :a.,B.,d.
Let the currents be as shown in
the figure.
Applying KVL along ABCDA,
we get
`-10i-2 + (2-i) 1 = 0`
`:. i = 0 `
Potential difference across = `(2-i)1 = 2XX1 = 2V` .


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