1. All the relevant resistors are in series, so the total (or equivalent) resistance is the sum of the resistances of the resistors: 20 Ω + 80 Ω + 50 Ω = 150 Ω [choice A].
2. The ammeter will read the current flowing through this circuit. We can find the ammeter reading using Ohm's law in terms of the electromotive force provided by the battery: I = ℰ/R = (30 V)(150 Ω) = 0.20 A [choice C].
3. The voltmeter will measure the potential drop across the 50 Ω resistor, i.e., the voltage at that resistor. We know from question 2 that the current flowing through the resistor is 0.20 A. So, from Ohm's law, V = IR = (0.20 A)(50 Ω) = 10. V, which will be the voltmeter reading [choice F].
4. Trick question? If the circuit becomes open, then no current will flow. Moreover, even if the voltmeter were kept as element of the circuit, voltmeters generally have a very high resistance (an ideal voltmeter has infinite resistance), so the current moving through the circuit will be negligible if not nil. In any case, the ammeter reading would be 0 A [choice B].
Answer:
a . 0.35cm
b. 11.33cm
Explanation:
a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x
#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:
Hence, for currents in same direction, the point is 0.35cm
b. Given both currents flow in opposite directions, the null point lies on the other side.
#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:
Let x be distance of N from first wire, then distance from 2nd wire is 4+x:
Hence, if currents are in opposite directions the point on x-axis is 11.33cm
1. electrons
2. positive to negative
3. insulator
4. TRUE
5. closed circuit
6. TRUE
7. series
8. TRUE
9. v=ir
10. TRUE
Hope this helps! :)
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