When one resistor in a parallel circuit is open-circuited, it effectively becomes an infinite resistance and no current flows through it. The total resistance of the parallel circuit increases, but the remaining resistors continue to function normally. The overall current through the circuit will decrease because the total current is now only dependent on the remaining active resistors. The voltage across all parallel resistors remains the same as before the open circuit occurred.
This happens because the total parallel resistance is lower than the individual resistors that make up the group of parallel resistors. When you add another parallel load, the resistance of that parallel group lowers and as result increases the current for the rest of the circuit.
When additional resistors are added in parallel to existing resistors in a circuit, the effective resistance decreases. This is because parallel resistors provide multiple pathways for current to flow, which reduces the overall resistance. The total or equivalent resistance ( R_{eq} ) can be calculated using the formula ( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} ), indicating that adding more parallel resistors results in a lower ( R_{eq} ).
If a short occurs in a resistor in series with other resistors, the voltage drops across the other resistors will increase. If a short occurs in a resistor in parallel with other resistors, the voltage drops across the other resistors will decrease, to zero.
Nothing happens.
The net resistance can be found out using the algebraic sums f series and parallel connections. When there is no current flowing in the circuit the net resistance is infinite.
The supply voltage in a parallel circuit remains the same regardless of the number of additional resistors connected. The voltage across each resistor in a parallel circuit is the same as the supply voltage. Adding more resistors in parallel will increase the total current drawn from the supply.
What do you mean by a 'parallel delta' circuit -is there such a connection.
This happens because the total parallel resistance is lower than the individual resistors that make up the group of parallel resistors. When you add another parallel load, the resistance of that parallel group lowers and as result increases the current for the rest of the circuit.
When additional resistors are added in parallel to existing resistors in a circuit, the effective resistance decreases. This is because parallel resistors provide multiple pathways for current to flow, which reduces the overall resistance. The total or equivalent resistance ( R_{eq} ) can be calculated using the formula ( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} ), indicating that adding more parallel resistors results in a lower ( R_{eq} ).
If a short occurs in a resistor in series with other resistors, the voltage drops across the other resistors will increase. If a short occurs in a resistor in parallel with other resistors, the voltage drops across the other resistors will decrease, to zero.
adding resistors to a string will have the effect of decreasing each resistor voltage drop.
Nothing
Current decreasesWhen voltage remains constant and resistance increases the current in the circuit will reduce.More informationV=IRwhere V is voltage,I is current andR is resistance.From the above equation,R=V/I, and hence resistance is indirectly proportional to current.Therefore, an increase in resistance would have the effect of decreased current.NB: this holds true only as long as the voltage remains constant.Another opinionHowever, this is only true in the case of a circuit connected in series.When circuits are connected in parallel, the opposite happens. If there is an increase in the amount of resistors in parallel, the total resistance of the circuit then decreases and the current increases subsequently.Yet another viewNo, that's not stated right.If more resistors are added in parallel - so that the circuit's overall total resistance decreases and its total current increases - that is NOT in any way the opposite of what this question is asking about...Let's make this crystal clear, so that there is no confusion: "an increase in the amount of resistors" is NOT the same as "an increase in resistance".So a parallel circuit behaves EXACTLY the same as a series circuit: if its overall resistance increases, the overall current going through the parallel circuit decreases AND if its overall resistance decreases, the overall current going through the parallel circuit increases.Actually, the second opinion is correctIn a parallel circuit, there are more branches to allow electrons back to the power supply, so current increases. With more resistors in a circuit, the overall resistance in a parallel circuit DECREASES.In a series circuit, current is the same throughout. So if more resistors are added, resistance INCREASES and so current DECREASES.
Nothing happens.
When two or more capacitors are connected in parallel across a potential difference, the total capacitance increases. This is because the equivalent capacitance of capacitors in parallel is the sum of their individual capacitances.
SMOKE!!! Yiii-haaaa! An ammeter is always as low a resistance as possible. That way, the current that you measure in a circuit using the meter will be the same as if the meter were not there. So putting an ammeter in parallel with a circuit means that you just short circuited the circuit you were intending to measure. Poof! There goes your ammeter!An ammeter has a very low internal resistance. So, if it is connected in parallel with a load, it will short-out that load. The resulting high current flowing through the ammeter may severely damage the meter (and possible harm the user), although most are fitted with fuses to protect them.
The net resistance can be found out using the algebraic sums f series and parallel connections. When there is no current flowing in the circuit the net resistance is infinite.