We use Kirchhoff's current and voltage laws to solve for the unknown currents.
The first equation is given by:
\[ I_1 + I_3 - I_2 = -2 ag{1} \]
The second equation is:
\[ I_3 + 2I_2 = 5 ag{2} \]
The third equation is:
\[ 2I_2 - (I_3 - I_2) - (I_1 + I_3 - I_2) = 5 ag{3} \]
From equation (3):
\[ I_1 = -\frac{11}{5} \, \text{A} \]
Therefore:
\[ y = 11 \]
\(y = 11\)
Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
In the figure shown below, a resistance of 150.4 $ \Omega $ is connected in series to an ammeter A of resistance 240 $ \Omega $. A shunt resistance of 10 $ \Omega $ is connected in parallel with the ammeter. The reading of the ammeter is ______ mA.
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is _____. 