Questions: R3. The battery supplies a constant voltage Vb, and the resistors are labeled with their resistances. The ammeters are ideal meters that read I1 and I2 respectively. The direction of each loop and the direction of each current arrow that you draw on your own circuits are arbitrary. Just assign voltage drops consistently and sum both voltage drops and currents algebraically and you will get correct equations. If the actual current is in the opposite direction from your current arrow, your answer for that current will be negative. The direction of any loop is even less important The equation obtained from a counterclockwise loop is the same as that from a clockwise loop except for a negative sign in front of every term (i.e., an inconsequential change in overall sign of the equation because it equals zero). Apply the junction rule to the junction labeled with the number 1 (at the bottom of the resistor of resistance R2). Answer in terms of given quantities, together with the meter readings I1 and I2 and the current I3. Sigma I=0=I3+I2-I1 You have indicated that I1 and I2 flow into the junction and I3 flows out. Is this what the diagram shows? If you apply the junction rule to the junction above R2, you should find that the expression you get is equivalent to what you just obtained for the junction labeled 1. Obviously the conservation of charge or current flow enforces the same relationship among the currents when they separate as when they recombine. Apply the loop rule to loop 2 (the smaller loop on the right). Sum the voltage changes across each circuit element around this loop going in the direction of the arrow. Remember that the current meter is ideal. Express the voltage drops in terms of Vb, I2, I3, the given resistances, and any other given quantities.

Transcript text: labeled $\mathrm{R}_{3}$. The battery supplies a constant voltage $V_{\mathrm{b}}$, and the resistors are labeled with their resistances. The ammeters are ideal meters that read $I_{1}$ and $I_{2}$ respectively. The direction of each loop and the direction of each current arrow that you draw on your own circuits are arbitrary. Just assign voltage drops consistently and sum both voltage drops and currents algebraically and you will get correct equations. If the actual current is in the opposite direction from your current arrow, your answer for that current will be negative. The direction of any loop is even less imporant The equation obtained from a counterclockwise loop is the same as that from a clockwise loop except for a negative sign in front of every term (i.e., an inconsequential change in overall sign of the equation because it equals zero). Figure 1 of 1 Part B Apply the junction rule to the junction labeled with the number 1 (at the bottom of the resistor of resistance $R_{2}$ ). Answer in terms of given quantities, together with the meter readings $I_{1}$ and $I_{2}$ and the current $I_{3}$. View Available Hint(s) \[ \Sigma I=0=I_{3}+I_{2}-I_{1} \] Subrit Previous Answers All attempts used; correct answer displayed You have indicated that $I_{1}$ and $I_{2}$ flow into the junction and $I_{3}$ flows out. Is this what the diagram shows? If you apply the juncion rule to the junction above $\boldsymbol{R}_{2}$, you should find that the expression you get is equivalent to what you just obtained for the junction labeled 1 . Obviously the conservation of charge or current flow enforces the same relationship among the currents when they separate as when they recombine. Part C Apply the loop rule to loop 2 (the smaller loop on the right). Sum the voltage changes across each circuit element around this loop going in the direction of the arrow. Remember that the current meter is ideal. Express the voltage drops in terms of $V_{\mathrm{b}}, I_{2}, I_{3}$, the given resistances, and any other given quantities. View Available Hint(s) $\square$ Submit