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Two cells are connected in opposition as shown. Cell E1 is of 8 V emf and 2 internal resistance; the cell E2 is of 2 V emf and 4 internal resistance. The terminal potential difference of cell E2 is V. [figure]
6
[figure]
I= E1-E2r1+r2= 8-22+4=1 A
E2 is being charged (current enters its + terminal)
V2=E2+Ir2=2+1 4=6 V
Answer: 6
When a potential difference V is applied across a wire of resistance R, it dissipates energy at a rate W. If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the energy dissipation rate will become:
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In the given circuit, the current in resistance R3 is : [figure]
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In the given figure, the charge stored in 6 capacitor, when points A and B are joined by a connecting wire is . [figure]
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A wire of resistance R and length L is cut into 5 equal parts. If these parts are joined parallely, then resultant resistance will be :
Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across A and B, their readings are V1, V2 and V3. Choose the correct option. [figure]
Wheatstone bridge principle is used to measure the specific resistance (S1) of given wire, having length L, radius r. If X is the resistance of wire, then specific resistance is ; S1=X ( r2L ). If the length of the wire gets doubled then the value of specific resistance will be :