Consider the circuit shown in the figure below, where R = 6.40 Ω and ΔV = 26.0 V. (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)
A circuit consists of two adjacent square loops, one to the right of the other. Starting from the lower left corner of each loop and going clockwise, points b, a, c, and d are at the corners of the left loop; points d, c, e, and f are at the corners of the right loop. A wire segment along a diagonal of the right loop connects points d and e.
The left side of the left loop, b a, contains a battery ΔV. The positive terminal is above the negative terminal.
The top side of the left loop, a c, contains a resistor R. Current I1 flows to the right from a to c.
The top side of the right loop, c e, contains a resistor of resistance 2.40 Ω. Current I3 flows to the right from c to e.
In the right side of the right loop, current I5 flows downward from e to f.
The bottom side of the right loop, f d, contains a resistor of resistance 9.00 Ω.
The bottom side of the left loop, d b, contains a resistor R.
The right side of the left loop, c d, which is the left side of the right loop, contains a resistor R. Current I2 flows downward from c to d.
The wire segment along the diagonal of the right loop, e d, which is the shared side between the triangular loops, contains a resistor R. Current I4 flows down and to the left from e to d.
Find each current (in A) indicated in the figure above. I1, I2, I3, I4, I5. (c)
Find the potential difference (in V) across each resistor.ΔVac, ΔVce, ΔVcd, ΔVed, ΔVfd, ΔVdb.
Find the power (in W) delivered to each resistor.
Pac, Pce, Pcd, Ped, Pfd, Pdb