What is the reading voltmeter at:
A to B
A to C
A to D
What formula can be used to calculate this?
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You calculate the voltage drops across the resistors caused by the 8 A current, as per Ohm's Law. Take proper note of the voltage's polarity: in a resistor the current flows from the higher voltage to the lower. Then add all the voltages, again keeping the polarity in mind. Decide which probe is the negative, for instance the left one. Then going from left to right you add a voltage if you encounter the |
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Firstly, you have to decide what to about the internal resistance of the batteries: probably, you can pretend that they are perfect voltage sources and so have zero series resistance. To determine the voltage at all the nodes, first pick some random number to be the voltage at point A. Since the choice of voltage doesn't matter, we might as well make it zero. So let us agree that A is at 0V. Next, to determine the voltages at all nodes (component-to-component connection points) we place a marker on the left and the move this marker across the components from node to node, making a note of the voltage changes. When we move the marker from A to B, the voltage drops because we are moving across a resistor, in the direction of the current ("flowing down the mountain, losing energy"). The voltage drops by 8 * 2 (V = IR). So at point B, we are at -16V relative to A = 0V. Next, our marker crosses a battery. The battery is such that the crossing starts at the minus terminal (short bar) ending at the positive (long bar). This means we are climbing the voltage hill: we are picking up 36V. So at point C, we are at -16 + 36 = 20V. (Remember, we agreed that batteries have no resistance and so the current doesn't factor in: we are assuming that a battery is just a pure voltage. This won't be true of a real battery!) In this manner, you can continue and determine the voltage at every node. Once you know the voltage at every node, knowing the potential difference between any two nodes is trivial: just subtract the first point from the second. A to C is 20 volts (20 - 0 = 20), and so on. Don't get tripped up by the second battery: it is in the opposite direction, so it gives us a voltage drop when we go left to right. You can make a line graph where on the horizontal axis you mark the nodes: A, B, C ... and on the vertical you show the voltages. It will look a little bit like a mountain range. Kirkhoff's Voltage Law says that if we trace these voltage differences around a complete circuit, when we return to the starting point, the voltage will agree: all of plus and minus voltage differences around the circuit cancel out, adding to zero. You don't have a complete circuit, so Kirkhoff's Voltage Law does not apply, but the process of tracing voltages will tell us the potential difference from A to D. Gravitational potential works similarly. If you make round trip around a landscape, returning to the starting point, you have no net gain or loss in gravitational potential energy, regardless of the elevation gains and drops you encountered along the way. Gravitational and electric fields are conservative. |
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Assuming arrow direction of current flow comes "out" of positive side on left... and assuming A to B means connecting + and - across A & B respectively...
If the person made the opposite assumptions to either convention, then you get a totally different answers which is why it is important to be consistent on this theoretical question. The reality of direction of physical direction of motion of free electrons in a conductor becomes irrelevant. It's just a standard notation for (analog) logic diagrams a.k.a. schematics and Theory of Operation descriptions. Vad could incorrectly become ... Vad= - 8 *( 2 + 1 + 1.5 ) -36V + 4V = -70V .. if you forget this standard notation. |
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