???? ONLINE CLASS LINK: https://www.cursoprojetoeletrico.com/ ⚡️ FREE CLASS: How to Make a Complete Electrical Installation from Scratch, Easily, Even if You Are Not a Professional Electrician: ➽ https://eletr.co/aula?utm_content=yt1... ------ ???? FAAALA ENGEHALL COMMUNITY ???? Join our community on WhatsApp to receive valuable tips and news in the electrical area, important information and much more. ???? ???? https://eletr.co/fala-eng?utm_content= ------ Did you like this video? ???? ✔️ Leave your comment and share with your friends ✔️ Follow our networks ➽ https://eletr.co/links - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Does a 7500W shower with a 6mm2 cable go bad? After all: What is the correct cable for me to use in a 7500W shower? FIRST STEP - We need to know what current this shower will consume. On average, a 7500W shower will always be connected to 220V, since the maximum power found for 127V showers is 5500W, that already comes from the factory, folks. So we already know the power (7500W) and we also know the mains voltage (220V). Now let's consult the Ohm's law table. We want to know the current. And since we have power and voltage, just replace P and V in the formula. So we have 7500W divided by 220V, which gives us a current of 34A. Now that we have found the current, let's move on to STEP TWO. Let's look at table 36 of NBR 5410 to check which cable gauge can handle this current. This table is for PVC insulated cables and we will follow the column of installation method B1, which refers to cables embedded in masonry. As we can see, in installation method B1 with 2 loaded conductors (after all, it is 220V, it can be phase and phase or phase and neutral), the recommended cable to be used that supports 34A is the 6mm2 cable, since the 4mm2 cable supports 32A. "There you go, André, so you're right! A 6mm² cable can handle a 7500W shower." That's not quite right, my dear! So far, we've only seen 2 points that we should pay attention to when choosing the right cable: The circuit current and the type of cable insulation. But these are not the only points that we should take into consideration. There are two other situations that are left out: one is the ambient temperature and the other is the number of circuit groups. A common mistake is to think that the ambient temperature is the temperature of the city where the installation will be carried out. This is not the case! The reference ambient temperature of 30°C that we find in the standard tables refers to the temperature of the environment where the installation is being carried out, as we can see in item 6.2.5.3.1 of NBR 5410. This temperature correction factor, or FCT, is most commonly used for industries and businesses, where the temperature is usually high (for example in bakeries) or low (for example in cold storage facilities). For homes, this factor will be unitary. But if you have to use this factor, just consult table 40 of NBR 5410. It has the various values ​​that we should adopt when the temperature is different from 30°. If the shower is installed in a bakery at 40º, for example, we need to multiply the current that the cable supports by a factor of 0.87 to make this correction, that is, it loses some of its current carrying capacity due to the increase in ambient temperature. Now let's consider the Grouping Factor, or FCA. Every time we have more than one circuit in a conduit, we must apply this factor. Hey André, but nowadays we put the shower circuit alone in an exclusive conduit, straight from the QDC to the shower! Today, yes, but in the past, no. Many old installations have a lot of cable squeezed inside the ducts and this causes damage. Take a look: Now let's imagine that, together with the shower circuit, 2 other circuits are coming out of the QDC in the same duct, so we have: Circuit 1: Lighting - 1.5mm2 cable Circuit 2: Sockets - 2.5mm2 cable Circuit 3: Shower - 6.0mm2 cable Let's consult table 42 of the standard. Our installation method is in a closed conduit (B1) and as we saw above, we have 3 grouped circuits. Note that we will use the grouping factor of 0.7. So we have a FCT equal to 1 and a FCA equal to 0.7. So let's go back to the calculations to find out the corrected design current, we use the following formula: Ic = Ip/(FCTxFCA) Corrected current = design current divided by the temperature correction factor times the grouping correction factor, thus: 34/(1x0.7) 34/(0.7) Ic= 49A