In the circuit shown epsilon_(1) and epsilon_(2) are two ideal sources of unknown emf. Some currents are shown in some branches of the circuit. Potential difference appearing across resistance 6Omega is V_(A)-V_(B)=10V. Then

In the circuit shown epsilon_(1) and epsilon_(2) are two ideal sources

1.	In the circuit shown epsilon_(1) and epsilon_(2) are two ideal sources of unknown emf. Some currents are shown in some branches of the circuit. Potential difference appearing across resistance 6Omega is V_(A)-V_(B)=10V. Then
Answer» The current in the CD branch is 5 AMP The UNKNOWN emf `e_(1)` is 40 volts The unknown emf `epsilon_(2)` is 68 volts The unknown resistance `R` is `9Omega` Solution :By APPLYING Kirchhof's law `epsilon_(1)=40V` `epsilon_(2)=68V` `R=9Omega` and current through `4Omega=5` amp

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In the circuit shown epsilon_(1) and epsilon_(2) are two ideal sources of unknown emf. Some currents are shown in some branches of the circuit. Potential difference appearing across resistance 6Omega is V_(A)-V_(B)=10V. Then

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