Answer to Question #55728 in Electric Circuits for Ojimi
a 2 ohms and 3 ohms resistors in parallel are connected in series to a 4 ohms resistor. the combination is then connected to a 12v battery, having internal resistance of 1 ohms. Calculate the current in the 3 ohms resistor
Given: R1=2 ohm, R2=3 ohm, R3=4 ohm, r=1 ohm, E=12V. Determine: I2=? Solution: First of all we have to find an effective resistance.Reff=R3+R1*R2/(R1+R2)=5.2 ohm. According to Ohm's law I=E/(r+R)=1.935 A. According to 1-st Kirchhoff rule I1+I2=I, from where I1=I-I2. In parallel connected resistors voltage drop is the same U1=U2, and by using Ohm's law I1*R1=I2*R2. Using I1=I-I2 we revrite (I-I2)*R1=I2*R2 and from this equation we determine I2=I*R1/(R1+R2)=0.774 A. The answer is I2=0.774 A.