A) \[\text{11}\text{.4 V and 11}\text{.5 V}\]
B) \[\text{11}\text{.7 V and 11}\text{.8 V}\]
C) \[\text{11}\text{.6 V and 11}\text{.7 V}\]
D) \[\text{11}\text{.5 V and 11}\text{.6 V}\]
Correct Answer: D
Solution :
[d] |
\[{{E}_{eq}}=\frac{\frac{12}{1}+\frac{13}{2}}{\frac{1}{1}+\frac{1}{2}}=\frac{37}{3}\] |
\[\frac{1}{{{r}_{eq}}}=\frac{1}{1}+\frac{1}{2}=\frac{3}{2}\Rightarrow {{r}_{eq}}=\frac{2}{3}\] |
\[I=\frac{\frac{37}{3}}{\frac{2}{3}+10}=\frac{37}{32}\] |
Voltage across the load\[=\left( \frac{37}{32} \right)(10)=\frac{370}{32}=11.56\,\,volt\] |
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