26.28 In the circuit shown in Fig. E26.28, find (a) the current in each branch and (b) the potential difference Vab of point a relative to point b. 26.29 The 10.00-V battery in Fig. E26.28 is removed from the cir- cuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the cir- cuit is as shown in the figure. Find (a) the current in each branch and (b) the potential difference Vab of point a relative to point b. .. Figure E26.28 2.00 2 10.00 V a www 1.00 2 5.00 V 10.00 Ω b 3.00 Ω 4.00 Ω

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Salve 26.29 ple

Figure E26.28
2.00 2 10.00 V
a
26.28 In the circuit shown in
Fig. E26.28, find (a) the current in each
branch and (b) the potential difference
Vab of point a relative to point b.
26.29 The 10.00-V battery in
Fig. E26.28 is removed from the cir-
cuit and reinserted with the opposite
polarity, so that its positive terminal is
now next to point a. The rest of the cir-
cuit is as shown in the figure. Find (a) the current in each branch
and (b) the potential difference Vab of point a relative to point b.
www
1.00 2 5.00 V
10.00 Ω
3.00 Ω
4.00 Ω
b
Transcribed Image Text:Figure E26.28 2.00 2 10.00 V a 26.28 In the circuit shown in Fig. E26.28, find (a) the current in each branch and (b) the potential difference Vab of point a relative to point b. 26.29 The 10.00-V battery in Fig. E26.28 is removed from the cir- cuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the cir- cuit is as shown in the figure. Find (a) the current in each branch and (b) the potential difference Vab of point a relative to point b. www 1.00 2 5.00 V 10.00 Ω 3.00 Ω 4.00 Ω b
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