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Answer :
Net resistance in the circuit(R) = Sum of the resistances of all resistors
R= 5 \(\Omega \)+ 8 \(\Omega \) + 12 \(\Omega \) = 25\(\Omega \)
We have;
\( R=\frac{V}I\)
\(25 \Omega =\frac{6V}I\)
or \(I=\frac{6V}{25 \Omega }=0.24A\)
Electric current remains the same through all resistors because resistances are connected in series.
Here we have,
Electric current, \(I\) = 0.24A
Resistance, R = 12\( \Omega \)
So, potential difference, V through the resistor of 12 \(\Omega \) = \(I\) x R
Or, V = 0.24A x 12 \(\Omega \) = 2.88 V
Thus, reading of ammeter = 0.24A
Reading of voltmeter through resistor of 12 \(\Omega \) = 2.88V
And the circuit can be formed as: