we have this 1/Re = (1/R1)+(1/R2) for calculating the effective resistance when resistances are connected in parallel
so the answer would be Re =( 70*30) / (70+ 30)
= 21 ohms
70 ohms and 30 ohms in parallel have a combined effective resistance of
( 1/70 + 1/30 )-1 = (70 x 30)/(70 + 30) = 2100/100 = 21 ohms
-- If the 20 ohms and the nother 10 ohms are configured in series, then the totalnet effective resistance is 30 ohms.-- If they are configured in parallel, then the total effective resistance is 62/3 ohms.
What would the measured ohms be for two 100 ohm resistors wired in series? Two 100 ohm resistors wired in series measure 200 ohms.
The net effective resistance of the parallel devices is the reciprocal of (1/12 + 1/4). Hence 3 ohms.=============================================(Which actually looks strangely similar to the first answer above.Could it be just coincidence ? I wonder . . . )
Well, the total circuit resistance depends on the type of connection. If the two resistances (or any number of resistances) are connected in series, IE. one resistance end is connected to one end of another resistance, the the circuit total resistance is the sum of the two resistances. say two resistances r1 and r2 are connected in series the total resistance is r1+r2 (in this case its 30 ohms). If the resistances are connected in parallel IE. both the ends of a resistance are connected to both ends of another resistance then the total resistance in this case shall be (r1*r2)/(r1+r2) ,( that is 6.67 ohms in given case).
In series like so ---6 ohms ---- 12 ohms --- , the total resistance is just 6 ohms + 12 ohms.assuming you mean in parallel like this:_|---6 ohms-----|-|~|-_|---12 ohms---|then the resistance of this can be calculated like so:1/6 ohms + 1/12 ohms = 1/R (where R is the resistance of the circuit as a whole)2/12 ohms + 1/12 ohms = 1/R3/12 ohms = 1/R1/4 ohms = 1/Rso R = 4 ohmsA few notes, if the resistors are in parallel the total resistance will always be less than or equal to the lowest resistance in parallel (i.e 6 ohms in parallel with 12 ohms will have resistance less than 6 ohms).Also if two resistances in parallel are the same, then the resistance is half of the resistance of both resistors (i.e. 1/2 ohms + 1/2 ohms = 1/R; 1 = 1/R, R=1 ohm which is half of 2 ohms).This process can be extended to 2 or more resistors in parallel.i.e if we had a 6 ohm, 6 ohm and 12 ohm resistor in parallel we could go1/6 ohms + 1/6 ohms + 1/12 ohms = 1/R(1/6 ohms + 1/6 ohms) + 1/12 ohms = 1/R1/3 ohms + 1/12 ohms = 1/R4/12ohms + 1/12 ohms = 1/R5/12 ohms = 1/Rso R = 12/5 ohms or 2.4 ohms
5 ohms
5 ohms.
The equivalent resistance of 75 ohms and 150 ohms in parallel is(75 x 150)/(75 + 150) = 50 ohms
5 ohms
The total resistance is 5 ohms. Scroll down to related links and look at "Parallel Resistance Calculator".
You can consider a short circuit to be a resistor with R=0 Ohms. It is then clear by the equation for calculation of parallel resistance that the combined resistance of a resistor in parallel to a short circuit is 0. Consider the following example with R1= 1k Ohms and R2= 0 Ohms: Rtotal = R1*R2 / (R1+R2) = R1*0 / R1 = 0 Ohms.
Three resistors in parallel: 20 ohms, 20 ohms, 10 ohms.1/ total resistance = (1/10) + (1/20) + (1/20) = (2/20) + (1/20) + (1/20) = 4/20 = 1/5 mho.Total resistance = 5 ohms
.9 watts.
The equivalent resistance of four resistors in parallel, the resistors being 2, 4, 6, and 8 ohms, is 0.96 ohms. RP = 1 / summationI=1toN (1/RI)
The power will be the product of the square of the current and the resistance of the load. The fact that the circuit is a parallel circuit is irrelevant to this question.P = I2R = 0.032 x 1000 =0.9 W
-- If the 20 ohms and the nother 10 ohms are configured in series, then the totalnet effective resistance is 30 ohms.-- If they are configured in parallel, then the total effective resistance is 62/3 ohms.
500 ohms.