P = E²/R
0.94 = E²/33
E = 5.56 volts
W = I^2 R
So, 0.94 = I^2 * 33
I = 168 mA
Other two in parallel are equal to 68*85/(68+85) = 37.7
ohms
Voltage across the two is 168mA x 37.7 ohms = 6.33 volts
Battery = 5.56 + 6.33 = 11.89 volts
Whats is the battery volatage? In the circuit shown in (Figure 1) the 33-Ohm resistor dissipates...
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