2.Determine the magnitude and sense (direction) of the current in the 20-Ω resistor when R= 15 **Figure 1: Circuit Diagram**This diagram represents an electrical circuit consisting of several key components:1. **Voltage Sources**: - A 30V battery is connected with its positive terminal on the left and negative on the right. - A 50V battery is connected in series, positioned above the resistors with its positive terminal on the top and negative at the bottom.2. **Resistors**: - There is a 10Ω resistor located at the bottom left of the circuit. - A 20Ω resistor is placed vertically adjacent to the 50V battery, with an upwards current labeled as "I". - Another resistor, labeled R, is on the right side of the circuit.3. **Current Flow**: - The current "I" flows upwards through the 20Ω resistor.This circuit can be analyzed to determine values such as the total resistance, current "I", and the voltage across each component using methods like Kirchhoff's laws or Ohm's Law.

Following figure showing the direction of the different currents in the circuit - Where, I2=I+I1Using the Kirchhoff's Voltage law in the loop ABEFA - 50 V-20I+10I1-30=0I1=-2 +2I .....(1) Using the Kirchhoff's Voltage law in the loop BCDEB - 15Ω×I2+20×I-50=015I2=50-20II2=10-4I3 .....(2) As we know- I2=I+I1Placing values from equation (1) and (2)10-4I3=I+(-2+2I)10-4I3=3I-210-4I=9I-613I=16I=1613I=1.23 A Magnitude of the current in the 20-Ω resistor is 1.23 A Given direction is correct in the circuit i.e. from bottom to top.