The current through the 40Ω resistor in the given circuit is 20mA, which is explained below.
Current through the resistors:
The volate across resistor R is 1V, as indicated by the voltmeter.
Since the potential of the battery is 3V, according to Kirchoff's voltage law, the voltage drop across 20Ω resistors must be 1V each since they are indentical.
Let a total current I passes through 20Ω resistor as it is connected to the battery in series. From Ohm's law:
20Ω × I = 1V
I = 50mA
Now the current indicated by the ammeter is 20mA which splits according to the reistors of 15Ω and 30Ω which are connected in parallel. Then again the current adds up to 20mA when leaving the set of parallel resistors, through the 40Ω resistor.
So the current through the 40Ω resistor is 20mA
Learn more about Kirchoff's voltage law:
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