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Two resistance R1=10 ohm and R2 =20 ohm are connected in series with a battery of emf E=10 vthen potential access resistance R1 isa) 3.333vb)6.666vc) 0.333vd) 0.666v |
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Answer» Mark it as BRAINLIEST Explanation: The circuit current I = 12/(10+20) I = 12/30 I =0.4 Amps. Power dissiparion in 10- OHM resistor P1= I^2R1 P1 = 0.4 X0. 4 x10 P1= 1.6 watts Power dissiparion in 20- ohm resistor P2= I^2R2 P2 = 0.4 x0. 4 x20 P2= 3.2 watts |
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