The ratio of number of turns of primary coil to secondary coil in a transformer is $2 : 3$ . If a source of $6 V$ ac is connected across the primary coil, then e.m.f. across the secondary coil (in volts) will be

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Solution

Given : number of turns in primary coil, $N_{p} = 2 N$ ,
number of turns in secondary coil $N_{s} = 3 N$ and
supply voltage on primary coil, $ϵ_{p} = 6 V$ .
Now, $\frac{ϵ_{P}}{ϵ_{S}} = \frac{N_{P}}{N_{S}}, ϵ_{S} = ϵ_{P} \times \frac{N_{S}}{N_{P}} = 6 \times \frac{3 N}{2 N} = 9 V .$

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The ratio of number of turns of primary coil to secondary coil in a transformer is 2:3. If a source of 6 V ac is connected across the primary coil, then e.m.f. across the secondary coil (in volts) will be

Given : number of turns in primary coil, Np=2N, number of turns in secondary coil Ns=3N and supply voltage on primary coil, ϵp=6V. Now, ϵPϵS=NPNS,ϵS=ϵP×NSNP=6×3N2N=9V.

The ratio of number of turns of primary coil to secondary coil in a transformer is $2 : 3$ . If a source of $6 V$ ac is connected across the primary coil, then e.m.f. across the secondary coil (in volts) will be

Given : number of turns in primary coil, $N_{p} = 2 N$ ,
number of turns in secondary coil $N_{s} = 3 N$ and
supply voltage on primary coil, $ϵ_{p} = 6 V$ .
Now, $\frac{ϵ_{P}}{ϵ_{S}} = \frac{N_{P}}{N_{S}}, ϵ_{S} = ϵ_{P} \times \frac{N_{S}}{N_{P}} = 6 \times \frac{3 N}{2 N} = 9 V .$