A few questions regarding electric circuits (calculations)?

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Q1 answer:

R(total) = 50 + (5 || 20) = 50 + 1/ (1/5 + 1/20) = 50 + 1/ (5/20) = 54 ohms

I (total) = V(battery) / R(total) = 12 / 54 = 2 / 9 Amp

i1 = current in 5 ohm i2 = current in 20 ohm
we have : i1 + i2 = 2/9 and i1 / i2 = 20 / 5 = 4
=> i1 = 8 / 45 amp and i2 = 2 / 45 amp
i1 ~ 0.178 amp and i2 ~ 0.044 amp


Q2 answer:

the exact value of the total current can be calculated as 3.182 amp.

if we consider the I(total)=3.2
we have V(voltmeter) = 3.2 * 10 ohm = 32 volts
=> V (40 ohm) = 50 - 32 = 18 volts
=> I (40 ohm) = 18 / 40 = 0.45 amp

with the more exact value of I (total) ~= 3.182
we have V(voltmeter) = 31.82 volts
=> V (40 ohm) = 50 - 31.82 = 18.18 volts
=> I (40 ohm) = 18.18 / 40 = 0.4545 amp



Q3 answer:

R(total) = 100 + (0.3 || 0.5) = 100 + ( 3 / 16 ) = 100.1875
I (ammeter) = 20 / 100.1875 = 0.19963 amp

c) just connect the voltmeter terminals to each end of the 0.3 resistor
the plus terminal to the top end (in diagram) and the minus terminal to the bottom end.

d) put the ammeter in series with the 0.3 resistor, i.e. draw it on the same wire as the 0.3
resistor

e) V = 20 - 100 * 0.19963 = 0.037 volt
I = 0.037 / 0.3 ~= 0.125 amp...Show more

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