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The switch in the circuit shown in Fig. has been closed for a long time. The switch opens at . Find for .
At first sight it's possible that we are scared about the transformation of the resistors of and . But it's completely unnecessary. Let's find the initials values of some variables in the circuit.
At , and behave like an open- and short-circuit respectively, so, will be in series with , and both are in parallel with , the equivalent resistance would be:
then, the current in that moment is:
Thus, is the voltage across the path containing the voltage source, the and the resistors. From the mesh of the left side of the source we may calculate the voltage drops in , that it's the same voltage that drops across and . Then we could find the current across , and then its voltage drops: Making an KVL in the mesh at left side of the source, we obtain:
then, if we applying Ohm's law:
However, we could obtain more directly by applying a voltage divisor between and , namely: .
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