There is a component that increases resistance and therefore reduces current, the resistor. I was wondering if there is a component I can use to increase current in a circuit.

# Electrical – component that increases current?

resistors

#### Related Solutions

we've got a capacitor and a series resistor and then point c

we've got a graph of "voltage at point c" that looks like it might be logarithmic.

from the text it can be understood that when voltage at point c drops low enough the deflector is activated. the deflector then operates autonomously for 2 seconds deploying and the retracts (presumably taking approximately another 2 seconds but that's not important)

we're told point C is connected to a voltage activated trigger

to have the deflector fully deployed at 3 seconds we need to delay for only 1 second as there are two seconds of delay inherent in the deflector. that means deploy at 6V

to increase the delay by mofiying the external components the easiest answer is to increase the capacitance of the capacitor, this will make the discharge take longer (I'm assuming the capacitor sis charged to a fixed voltage - this seems like a reasonable assumption)

the effect of the resistor on the voltage at point C is more complicated.

assuming the bottom end of the capacitor is held at a constant voltage during the discharge. We aren't told this, but from the graph it seems likely, (and we aren't given any other point to measure voltage from!) point C behaves somwwhat like a resistor to ground: the more current is flowing into it the higher the voltage.

looking carefully at the graph reveals that it is not exponetial decay. in the first second the voltage halves, but in the next second it doesn't. so point C isn't pure resistance to ground... looking even more closely, I am unable to fit any exponential curve to that graph, thus the terminal behind point C can't be treated as a fixed thevenin equivalent circuit.

The graph looks approimately hyperbolic, I can't immagine what circuitry or electrical behaviour behind terminal C would be required to get that effect.

Either this is a trick question, or it is very badly written.

But it is known that current through the diode will not increase that much(a tiny increase will happen).

Once the diode is turned on, a change in voltage across it will cause a **large** change in current, not a small one.

Isn't this a contradiction that elements connected in series shares the same current ?

No. The load line graph shows exactly what will happen. If you increase the source voltage, the green line will shift to the right. Then you will find a new intersection between the IV curve for the source+resistor and the diode.

It will not cause a large increase in the diode current because the resistor limits the current. This is because \$V_d = V_s - V_r\$. The diode voltage is equal to the source voltage minus the resistor voltage. So as the source voltage rises, either the diode or the resistor can take up that change. And because the diode has very low differential resistance, it's the resistor that will take up most of the added voltage.

**Edit**

Another way you know this is not a contradiction is that the fact the resistor and diode currents are equal was a fundamental part of drawing the load line graph.

The two curves are

$$i=I_s \exp \left(\frac{nv}{V_T}-1\right)$$

and

$$i=\frac{V_s-v}{R}$$

The \$v\$ on the x-axis is the voltage across the diode. The \$i\$ on the y-axis is the current flowing in common through all three elements in the circuit (source, resistor, and diode).

## Best Answer

resistors in parallel will do the job. less resistance and increasing the current.